复旦大学：《数学物理方法 Methods of Mathematical Physics》课程教学资源（教学研究）Transverse vibrations of a thin loaded rod：theory and experiment.pdf_大学文库

loP Publishing EuJ.Phys.36(2015)055035(11pp) Transverse vibrations of a thin loaded rod theory and experiment Jue Xu, Yuan-jie Chen and Yong-li Ma Department of Nuclear Science and Technology, Fudan University, Shanghai 200433 Peoples Republic of China artment of Physics, Fudan University, Shanghai 200433, Peoples Republic of E-mail:yIma@fudan.edu.cn Received 3 March 2015, revised 28 April 2015 Accepted for publication 16 June 2015 Published 4 August 2015 Abstract The general formulation of a determinate solution problem is deduced for the transverse vibrations of a thin loaded rod. The vibration frequencies of a thin homogeneous rod carrying a concentrated mass as a function of the loads position and mass are exactly solved. The dynamic measurement method of Youngs modulus of the rods is presented within this theory. Our measure- ments of Youngs modulus in the dynamic method agree with those in the traditional bending method, therefore the theory is verified by our experiments Keywords: transverse vibrations of a load rod, Youngs modulus, dynamic measurement method (Some figures may appear in colour only in the online journal) 1 Introduction In college-level instruction on vibration and waves, the model is usually simplified as vibrations of a single particle even for a tuning fork. Actually, the tuning fork consists of two elastic rods and we need to analyze the transverse vibrations of these rods. The textbooks on he method of mathematical physics have only the longitudinal vibrations of a rod [1-3] and rarely refer to its transverse vibrations [4-6] with and without a load. Even in engineering mathematics [4] and applied mathematics [5], the equation of motion on the vertical vibra- tions of a thin long rod is only directly given and its derivation is not present In the course of the method of mathematical physics, the transverse vibrations of a rod were first introduced by [6], but its deduction on the solving problem is too simple to follow by students. Even in transverse vibrations of a rod is still simple [7, 8] although its longitudinal vibration is 0143-0807/15/055035411$33.00 G 2015 IOP Publishing Ltd Printed in the UK

Transverse vibrations of a thin loaded rod: theory and experiment Jue Xu1 , Yuan-jie Chen2 and Yong-li Ma2 1 Department of Nuclear Science and Technology, Fudan University, Shanghai 200433, People’s Republic of China 2 Department of Physics, Fudan University, Shanghai 200433, People’s Republic of China E-mail: ylma@fudan.edu.cn Received 3 March 2015, revised 28 April 2015 Accepted for publication 16 June 2015 Published 4 August 2015 Abstract The general formulation of a determinate solution problem is deduced for the transverse vibrations of a thin loaded rod. The vibration frequencies of a thin homogeneous rod carrying a concentrated mass as a function of the load’s position and mass are exactly solved. The dynamic measurement method of Young’s modulus of the rods is presented within this theory. Our measurements of Young’s modulus in the dynamic method agree with those in the traditional bending method, therefore the theory is verified by our experiments. Keywords: transverse vibrations of a load rod, Youngʼs modulus, dynamic measurement method (Some figures may appear in colour only in the online journal) 1. Introduction In college-level instruction on vibration and waves, the model is usually simplified as vibrations of a single particle even for a tuning fork. Actually, the tuning fork consists of two elastic rods and we need to analyze the transverse vibrations of these rods. The textbooks on the method of mathematical physics have only the longitudinal vibrations of a rod [1–3] and rarely refer to its transverse vibrations [4–6] with and without a load. Even in engineering mathematics [4] and applied mathematics [5], the equation of motion on the vertical vibrations of a thin long rod is only directly given and its derivation is not present. In the course of the method of mathematical physics, the transverse vibrations of a rod were first introduced by [6], but its deduction on the solving problem is too simple to follow by students. Even in the mechanics of vibration, a fundamental course for a mechanics major, the instruction on transverse vibrations of a rod is still simple [7, 8] although its longitudinal vibration is European Journal of Physics Eur. J. Phys. 36 (2015) 055035 (11pp) doi:10.1088/0143-0807/36/5/055035 0143-0807/15/055035+11$33.00 © 2015 IOP Publishing Ltd Printed in the UK 1

2015)055035 J Xu et al discussed with a load [7]. Earlier papers have addressed the problem of vibrating rods [9, 10] In special textbooks [11], the exact solutions of the rods transverse vibrations are put forward systematically, but only a few solutions at special conditions are given [12, 13].It is necessary to derive a general formulation of the determinant solution problem for the transverse vibrations of a thin loaded rod, and find the exact solutions in general conditions Much work has been done on the more complicated case of anisotropic vibrations of rods. For example, the greatest use of piezoelectric resonators is the famous 32.768 kHz quartz tuning fork, including a review paper [14], the fabrication [15], and the analysis of frequency [16]. Although this work treats only the isotropic case, which is a limiting work in the mechanics major, we emphasize the physical foundation for undergraduates in the physics major Knowledge of Youngs modulus is fundamentally important to understand the mechanical behaviour of materials, such as metals, ceramic grinding stones, dental compo sites, and polymers [17-20). Youngs moduli are determined traditionally by the static and dynamic methods. In static measurements [21, 22], such as the classical tensile or com- pressive test, a uniaxial stress is exerted on the material, and the elastic modulus is calculated from the transverse and axial deformations as the slope of the stress-strain curve at the origin The static methods include the three-point bending [17-20], four-point bending [23], clamped beam, and compression/tension stress, etc. A dynamic method of measuring Youngs mod- ulus of stalloy was described in an earlier paper [24] using a loaded fixed-free bar vibrating in flexure and developed to measure Youngs modulus of elasticity of a solid [25], carrying a heavy mass of precise finite dimensions at the free end and giving the derivation of the quation of motion. Dynamic methods [26-29] are more precise since they use very small strains, far below the elastic limit and therefore are virtually nondestructive and allow repeated testing of the same sample. These include the ultrasonic pulse-echo [27] ,bar resonance methods [22, 28, 29], travelling or standing wave, bending/transversal or long itudinal wave, transient pulse generation, etc. Recently, a new vibration beam technique [30] for the determination of the dynamic Youngs modulus has been developed, but without the dded loads. The dynamic methods redeem the defects that the static bending method cannot be applied to the measurement of fragile materials. Our method has added a variable with the loads at different positions and different masses. For didactic purposes undergraduates majoring in physics, in this paper we derive a general formulation of the determinant solution problem for the transverse vibrations of a thin loaded rod, obtain an exact solution of the problem, and deduce a general relationship between eigenfrequencies and the loads position and mass. Different resonance frequencies are measured by adding both the same mass to different positions of the rod and different masses to the same position of the rod. We deal with Youngs modulus measurement method based on the vibration of a thin long rod with added point mass. The elastic modulus is calculated from the least square fit of frequency versus the square of the wave number calculated from the characteristic equation. According to this model, a new kind of dynamic measurement method of Youngs modulus is presented. This method is more comprehensive and is advanced when it is not convenient to change the length of samples The paper is organized as follows. The mechanical model and the solution for the of the load's position and 2.A practical implementation and the results of the experiments are described in section 3. showing results in agreement with model predictions. A summary is given in section 6

discussed with a load [7]. Earlier papers have addressed the problem of vibrating rods [9, 10]. In special textbooks [11], the exact solutions of the rod’s transverse vibrations are put forward systematically, but only a few solutions at special conditions are given [12, 13]. It is necessary to derive a general formulation of the determinant solution problem for the transverse vibrations of a thin loaded rod, and find the exact solutions in general conditions. Much work has been done on the more complicated case of anisotropic vibrations of rods. For example, the greatest use of piezoelectric resonators is the famous 32.768 kHz quartz tuning fork, including a review paper [14], the fabrication [15], and the analysis of frequency [16]. Although this work treats only the isotropic case, which is a limiting work in the mechanics major, we emphasize the physical foundation for undergraduates in the physics major. Knowledge of Young’s modulus is fundamentally important to understand the mechanical behaviour of materials, such as metals, ceramic grinding stones, dental composites, and polymers [17–20]. Young’s moduli are determined traditionally by the static and dynamic methods. In static measurements [21, 22], such as the classical tensile or compressive test, a uniaxial stress is exerted on the material, and the elastic modulus is calculated from the transverse and axial deformations as the slope of the stress-strain curve at the origin. The static methods include the three-point bending [17–20], four-point bending [23], clamped beam, and compression/tension stress, etc. A dynamic method of measuring Young’s modulus of stalloy was described in an earlier paper [24] using a loaded fixed-free bar vibrating in flexure and developed to measure Young’s modulus of elasticity of a solid [25], carrying a heavy mass of precise finite dimensions at the free end and giving the derivation of the equation of motion. Dynamic methods [26–29] are more precise since they use very small strains, far below the elastic limit and therefore are virtually nondestructive and allow repeated testing of the same sample. These include the ultrasonic pulse-echo [27], bar resonance methods [22, 28, 29], travelling or standing wave, bending/transversal or longitudinal wave, transient pulse generation, etc. Recently, a new vibration beam technique [30] for the determination of the dynamic Young’s modulus has been developed, but without the added loads. The dynamic methods redeem the defects that the static bending method cannot be applied to the measurement of fragile materials. Our method has added a variable with the loads at different positions and different masses. For didactic purposes undergraduates majoring in physics, in this paper we derive a general formulation of the determinant solution problem for the transverse vibrations of a thin loaded rod, obtain an exact solution of the problem, and deduce a general relationship between eigenfrequencies and the load’s position and mass. Different resonance frequencies are measured by adding both the same mass to different positions of the rod and different masses to the same position of the rod. We deal with Young’s modulus measurement method based on the vibration of a thin long rod with added point mass. The elastic modulus is calculated from the least square fit of frequency versus the square of the wave number calculated from the characteristic equation. According to this model, a new kind of dynamic measurement method of Young’s modulus is presented. This method is more comprehensive and is advanced when it is not convenient to change the length of samples. The paper is organized as follows. The mechanical model and the solution for the eigenfrequencies as a function of the load’s position and mass are illustrated in section 2. A practical implementation and the results of the experiments are described in section 3, showing results in agreement with model predictions. A summary is given in section 6. Eur. J. Phys. 36 (2015) 055035 J Xu et al 2

2015)055035 J Xu et al 2. The mechanical model and the solutions We consider a rod of length I along the x-axis in equilibrium. The mass of the rod is m=/dxp(x)s(r) with dx being the line element, P(x) being the volume density, and s(x)=w(r)h(x) being the cross-section area. Here the width of the rod is y= w(r)and the thickness is z= h(x). The turning radius of this cross-section r(x) satisfies zada h(x)Jo 12 Figure 1 shows the diagram and cross-section of the thin homogenous rod. Let the element be dm= p(x)dr with the volume element dr= s()dx, one obtains the rotate inertia to the ov-axis as h(x) When the rod deforms transversely, its every cross-section should produce shearing rces. Let the shearing force on the left of volume element be o(x, t)(down direction), and the right one be Q=o(x, t)+dQ(, t)dx(up direction) with d being an abbreviation of d/dx. These two shearing forces form a force couple, which bends the rod. Figure 2 shows an element of a thin homogenous rod in bending time point (x, D), is u(x, t). The curvature radius of the bending d"&s of a thin loaded rod We consider the characteristic quantity of the transverse vibration the displacement of the rod away from the equilibrium position along the z-direction at space When the rod bends, the central line length dr remains unchanged. However, the upper part of the central line that suffers the tension of the nearby elements is prolonged; the lower part that suffers the pressure is compressed. Consequently, the force couple consists of tension and pressure, the so-called bending moment. Let the bending moment on the left of the volume element be M(x, t)(clockwise), and the bending moment on the right be M=M(x, t)+ dM(r, t)dx(counter-clockwise). The bending moments act as resistance to the bending of the rod, and lead the system to the dynamic equilibrium states. Figure 3 shows the force acting on an element of a thin homogenous rod. As shown in figure 2, we take a lamina with thickness dz at z position and width w(x). We recall that the length is dx at the center line of the volume element The length of the lamina is (R+z)de= dx zdx/R and the relative extension is Z/R. So that the tensile stress is P=-Yz/R with y being Youngs modulus. The tension element is dG= Pwdz, the bending moment element is dM= zdG=-Y22wdz/R and the bending moment is M=-YJ/R withJ=/22wdz=s(x)r2(r) the inertia moment per mass for the cross-section s(x)to the For the mass element p(r)dr of the bending rod, the inertia force is -p(x)drude, the external force is f(x, t)dr, and the external bending moment is m(x, t)dx. The equilibrium equation of moment to the left center C is

2. The mechanical model and the solutions We consider a rod of length l along the x-axis in equilibrium. The mass of the rod is m x xsx = ∫ d () ( ρ ) l 0 with dx being the line element, ρ( ) x being the volume density, and s() () () x wxhx = being the cross-section area. Here the width of the rod is y wx = ( ) and the thickness is z = h x( ). The turning radius of this cross-section r(x) satisfies r x = = ∫ h x z z h x ( ) 2 ( ) d ( ) 12 . (1) h x 2 0 ( ) 2 1 2 2 Figure 1 shows the diagram and cross-section of the thin homogenous rod. Let the mass element be d ( )d m x = ρ τ with the volume element d ( )d τ = sx x, one obtains the rotational inertia to the oy-axis as I = ρ τ h x d x ( ) 12 yy ( )d . (2) When the rod deforms transversely, its every cross-section should produce shearing forces. Let the shearing force on the left of volume element be Q(, ) x t (down direction), and the right one be Q′= +∂ Qx t Qx t x ( , ) ( , )d x (up direction) with ∂x being an abbreviation of ∂ ∂x. These two shearing forces form a force couple, which bends the rod. Figure 2 shows an element of a thin homogenous rod in bending. We consider the characteristic quantity of the transverse vibrations of a thin loaded rod, the displacement of the rod away from the equilibrium position along the z-direction at spacetime point (x,t), is ux t (, ). The curvature radius of the bending is = +∂ ∂ ⎡ ⎣ ⎤ ⎦ R uu 1 . (3) ( ) x xx 2 3 2 2 When the rod bends, the central line length dx remains unchanged. However, the upper part of the central line that suffers the tension of the nearby elements is prolonged; the lower part that suffers the pressure is compressed. Consequently, the force couple consists of tension and pressure, the so-called bending moment. Let the bending moment on the left of the volume element be Mxt ( , ) (clockwise), and the bending moment on the right be M Mxt Mxt x ′= +∂ ( , ) ( , )d x (counter-clockwise). The bending moments act as resistance to the bending of the rod, and lead the system to the dynamic equilibrium states. Figure 3 shows the force acting on an element of a thin homogenous rod. As shown in figure 2, we take a lamina with thickness dz at z position and width w(x). We recall that the length is dx at the center line of the volume element. The length of the lamina is (R z x zx + =+ )d d d θ R and the relative extension is z R. So that the tensile stress is P Yz = − R with Y being Young’s modulus. The tension element is dG = Pwdz, the bending moment element is dd d M z G Yz w z = =− R 2 , and the bending moment is M YJ R = − , (4) with J zw z sxr x = = ∫ d () ( ) 2 2 the inertia moment per mass for the cross-section s(x) to the center oy‐axis. For the mass element ρ( )d x τ of the bending rod, the inertia force is − ∂ ρ τ () d x utt 2 , the external force is fxt ( , )dτ, and the external bending moment is mxt x ( , )d . The equilibrium equation of moment to the left center C is Eur. J. Phys. 36 (2015) 055035 J Xu et al 3

Eur.J.Phys.36(2015)055035 J Xu et al sin(k-k)cosh(k-k)+ sin k cosh k'cosh(k-k) sinh k cos k' cos(k-k) (24) From this equation we can obtain the eigenvalues kn as a function of x'/, m/m for n=0, 1, 2,.They determine the available frequencies of the rod In some special cases, equation(24)can be simplified. For example, it becomes equation(6) of [13] for c= oo(the end of x=l is hinged by a rotational spring of constant stiffness c; n our work, it is easy to include the finite c). It also becomes 1 +cos k cosh k k-(sin k cosh k- cos k sinh k) for i'=1. It can further be simplified as 1 cos k cosh k=0 for m'=0 without a load. In this case, Youngs modulus is Y =38.32 /pf/h after solving the fundamental eigenvalue ko with f= o/(2r). This is the theoretical basis of Youngs modulus measurement in the traditional dynamic method [17]. And now we extend it by adding two degrees of freedom of x'/l and m/m. The traditional dynamic method is only a special case of our model. 3. Experiments and results Experimental setup consists of the following apparatus: two thin homogenous rods, several magnet loads, an oscilloscope, two supports and bases, and a photometer. We do not draw the schematic diagram of the experimental apparatus and setups. The heavy base is used to fix the homogenous rod. The photometer records the time intervals of the rod blocking light and outputs the voltage signal to the oscilloscope. There is a linear light source on one side of the photometer and a photo-resistor on the other side. The light intensity changes the resistance of photometer, which changes the voltage of the photo-resistor. Without obstruction in front of he photometer, the light irradiates the photo-resistor to lower its resistance; otherwise, the resistance increases. Thus, the vibration frequencies of the rod can be read by observing the periods of the voltage signal In the experiments, we put two magnet loads at x' of the rod on the y-direction sym- metrically, and clap the endpoint of x= I to start the transverse vibrations. In the beginning, there is a fundamental frequency and higher harmonic frequencies. After a moment, the higher harmonic frequencies decay quickly, and the rod vibrates at the fundamental fre- quency. At this time, the fundamental frequency can be read out by observing the wave shape on the oscilloscope. By adding two magnet loads at the same position symmetrically, we measure the fundamental frequency and obtain the mass-dependent fundamental frequency at the fixed x'. For a given x, on the other hand, the eigenvalues k, which correspond to the different mass of loads, are obtained by solving equation (24) numerically. From equation( 25), it is clear that the fundamental frequency is proportional to k.Thus Young's modulus can be fitted by the least square method according to the linear relationship between requencies a and eigenvalues k2 In the similar approach, keeping the masses of loads and changing the positions of loads the eigenvalues k that correspond to the different positions of loads are solved from equation(24)numerically and Youngs modulus can be fit by the least square method

′ + = ′ ′− ′ ′+ − ′ − ′ − −′ −′+ ′ −′ − ′ −′ m m k k k k k k k kk kk kk kk k k kk k k kk 2 (1 cos cosh ) sin cosh cos sinh sinh( )cos( ) sin( )cosh( ) sin cosh cosh( ) sinh cos cos( ). (24) From this equation we can obtain the eigenvalues kn as a function of xlmm ′ ′ , for n = 0, 1, 2 ,.... They determine the available frequencies of the rod ω ρ = ′ ′ ⎜ ⎟ ⎛ ⎝ ⎞ ⎠ r l Y k x l m m n , . (25) n 2 2 In some special cases, equation (24) can be simplified. For example, it becomes equation (6) of [13] for c = ∞ (the end of x = l is hinged by a rotational spring of constant stiffness c; in our work, it is easy to include the finite c). It also becomes 1 cos cosh + = k k − ′ k (sin cosh cos sinh k k kk) m m for x¯′ = 1. It can further be simplified as 1 cos cosh 0 + = k k for m′ = 0 without a load. In this case, Young’s modulus is Y = 38.32 lf h ρ 422 after solving the fundamental eigenvalue k0 with f = ω π (2 ). This is the theoretical basis of Young’s modulus measurement in the traditional dynamic method [17]. And now we extend it by adding two degrees of freedom of x l ′ and m m′ . The traditional dynamic method is only a special case of our model. 3. Experiments and results Experimental setup consists of the following apparatus: two thin homogenous rods, several magnet loads, an oscilloscope, two supports and bases, and a photometer. We do not draw the schematic diagram of the experimental apparatus and setups. The heavy base is used to fix the homogenous rod. The photometer records the time intervals of the rod blocking light and outputs the voltage signal to the oscilloscope. There is a linear light source on one side of the photometer and a photo-resistor on the other side. The light intensity changes the resistance of photometer, which changes the voltage of the photo-resistor. Without obstruction in front of the photometer, the light irradiates the photo-resistor to lower its resistance; otherwise, the resistance increases. Thus, the vibration frequencies of the rod can be read by observing the periods of the voltage signal. In the experiments, we put two magnet loads at x′ of the rod on the y-direction symmetrically, and clap the endpoint of x = l to start the transverse vibrations. In the beginning, there is a fundamental frequency and higher harmonic frequencies. After a moment, the higher harmonic frequencies decay quickly, and the rod vibrates at the fundamental frequency. At this time, the fundamental frequency can be read out by observing the wave shape on the oscilloscope. By adding two magnet loads at the same position symmetrically, we measure the fundamental frequency and obtain the mass-dependent fundamental frequency at the fixed x′. For a given x′, on the other hand, the eigenvalues k, which correspond to the different mass of loads, are obtained by solving equation (24) numerically. From equation (25), it is clear that the fundamental frequency is proportional to k 2 . Thus Young’s modulus can be fitted by the least square method according to the linear relationship between frequencies ω and eigenvalues k 2 . In the similar approach, keeping the masses of loads and changing the positions of loads, the eigenvalues k that correspond to the different positions of loads are solved from equation (24) numerically and Young’s modulus can be fit by the least square method. Eur. J. Phys. 36 (2015) 055035 J Xu et al 7

Eur. J. Phy 2015)055035 J Xu et al Figure 4. The vibration frequencies f in the units of Hz versus the dimensionless eigenvalues k- with the different load masses for two kinds of thin homogenous Fe (red) and Cu(blue)rods. From equation(24), we can see that f= sk- with the slope s For two kinds of in homogeneous rods, iron(Fe)and copper(Cu), the experimental parameters are 1=0.26 m,h=1020×10-3m,r=2945×10-m,andp=790×103kgm3,andl=0.26m h= 1.027 x 10-3m, r=2.965 10-m, and p=8.34 x 10 kg m, respectively. Our observations for f and calculation on k versus the different masses of the loads are shown in figure 4. The red dots(blue squares)are the experimental data for the Fe( Cu)rod; while the red line(blue line) is the theoretical result for the Fe( Cu) rod. By using the least square method and data from figure 4, we obtain the slope SFe =(3.31+ 0.05)Hz and Youngs modulus YFe =(1.81+0.03)x 10 1 Nm- for the iron rod, and the slope cu=(249±0.05) Hz and young's modulus y=(1.6±0.03)×10Nm-2 for the copper rod. The speeds of sound for the shear wave v= vLp are vFe=4.787 x 10ms nd vCu=3.565 x 10 ms" in our experiments With a similar approach, our observations for f and calculation on k- versus the different ositions of the loads are shown in figure 5. We obtain SFe=(3. 27+ 0.05)Hz and he=(1.75±0.02)×1031Nm-2 for the iron rod, and So=(243±0.05) Hz and u=(100±001)×101Nm2 for the copper rod. To compare our method with other methods, we independently determine Youngs modulus of the above rods with the three-point bending method [17-20]. This is a traditional method in experimental physics that teaches measuring of Youngs modulus of materials. The displacement 4z at the position of loads x= 1/2 is linear with the total mass M'=m+M i.e., 4z=5M with the slope 5=4 wr, the load mass m and the tray mass M. Here d=0.23m is the distance between the support, g is the gravitational acceleration, and w=0.023m is the width of the rod Our observations for 4z as a function of m' are shown in figure 6. The red dots(blue squares)are the experimental data for the above Fe( Cu)rod, while the red line(blue line)is the linear fitting for the same Fe(Cu)rod. Here, s is the slope of the straight line and sM is its ordinate at the origin. From figure 6, we can fit the slopes 5fc=0.0066 m kg" and

From equation (24), we can see that f k = ζ 2 with the slope ζ = π ρ eq l 1 Y 2 2 . For two kinds of thin homogeneous rods, iron (Fe) and copper (Cu), the experimental parameters are l = 0.26 m, = × − h 1.020 10 3 m, = × − r 2.945 10 4 m, and ρ = × 7.90 103 kg m−3 , and l = 0.26 m, = × − h 1.027 10 3 m, = × − r 2.965 10 4m, and ρ = × 8.34 103 kg m−3 , respectively. Our observations for f and calculation on k 2 versus the different masses of the loads are shown in figure 4. The red dots (blue squares) are the experimental data for the Fe (Cu) rod; while the red line (blue line) is the theoretical result for the Fe (Cu) rod. By using the least square method and data from figure 4, we obtain the slope ζFe = ± (3.31 0.05)Hz and Young’s modulus YFe =±× (1.81 0.03) 1011 N m−2 for the iron rod, and the slope ζCu = ± (2.49 0.05)Hz and Young’s modulus YCu =±× (1.06 0.03) 1011 N m−2 for the copper rod. The speeds of sound for the shear wave v Y = ρ are vFe = × 4.787 103 m s−1 and vCu = × 3.565 103 m s−1 in our experiments. With a similar approach, our observations for f and calculation on k 2 versus the different positions of the loads are shown in figure 5. We obtain ζFe = ± (3.27 0.05)Hz and YFe =±× (1.75 0.02) 1011 N m−2 for the iron rod, and ζCu = ± (2.43 0.05) Hz and YCu =±× (1.00 0.01) 1011 N m−2 for the copper rod. To compare our method with other methods, we independently determine Young’s modulus of the above rods with the three-point bending method [17–20]. This is a traditional method in experimental physics that teaches measuring of Young’s modulus of materials. The displacement Δz at the position of loads x l = 2 is linear with the total mass MmM ′= ′+ , i.e., Δz M = ′ ξ with the slope ξ = d g 4 h wY 3 3 , the load mass m′ and the tray mass M. Here d = 0.23m is the distance between the support, g is the gravitational acceleration, and w = 0.023m is the width of the rod. Our observations for Δz as a function of m′ are shown in figure 6. The red dots (blue squares) are the experimental data for the above Fe (Cu) rod, while the red line (blue line) is the linear fitting for the same Fe (Cu) rod. Here, ξ is the slope of the straight line and ξM is its ordinate at the origin. From figure 6, we can fit the slopes ξFe = 0.0066 m kg−1 and Figure 4. The vibration frequencies f in the units of Hz versus the dimensionless eigenvalues k 2 with the different load masses for two kinds of thin homogenous Fe (red) and Cu (blue) rods. Eur. J. Phys. 36 (2015) 055035 J Xu et al 8