University Physics Al No 5 Simple Harmonic Oscillation Class Number ame L. Choose the Correct Answer 1. a particle on a spring executes simple harmonic motion. If the mass of the particle and the amplitude are both doubled then the period of oscillation will change by a factor of (D) (C)2 2 Since the period of oscillation is T===2,/, according to the problem Wkvk k 2. A particle on a spring executes simple harmonic motion; when it passes through the equilibrium position it has a speed v. The particle is stopped, and then the oscillations are restarted so that it now passes through the equilibrium position with a speed of 2v. After this change the frequency of oscillation will change by a factor of (E) (B)√8 Since the frequency of oscillation is v= 2r rvm the frequency of oscillation will not 3. A particle on a spring executes simple harmonic motion; when it passes through the equilibrium position it has a speed v. The particle is stopped, and then the oscillations are restarted so that it now passes through the equilibrium position with a speed of 2v. After this change the maximum displacement will change by a factor of (C) (A)4 (C)2 (D)√2 (E)1 Solution The energy of simple harmonic motion is E=kA Ek+e
University Physics AI No. 5 Simple Harmonic Oscillation Class Number Name I. Choose the Correct Answer 1. A particle on a spring executes simple harmonic motion. If the mass of the particle and the amplitude are both doubled then the period of oscillation will change by a factor of ( D ) (A) 4. (B) 8 . (C) 2. (D) 2 Solution: Since the period of oscillation is k m T π ω π 2 2 = = , according to the problem T k m k m k m T 2 2 2 2 2 = 2 = ⋅ = ′ ′ = π π π 2. A particle on a spring executes simple harmonic motion; when it passes through the equilibrium position it has a speed v. The particle is stopped, and then the oscillations are restarted so that it now passes through the equilibrium position with a speed of 2v. After this change the frequency of oscillation will change by a factor of ( E ) (A) 4. (B) 8 . (C) 2. (D) 2 (E) 1 Solution: Since the frequency of oscillation is m k π π ω ν 2 1 2 = = , the frequency of oscillation will not change. 3. A particle on a spring executes simple harmonic motion; when it passes through the equilibrium position it has a speed v. The particle is stopped, and then the oscillations are restarted so that it now passes through the equilibrium position with a speed of 2v. After this change the maximum displacement will change by a factor of ( C ) (A) 4. (B) 8 . (C) 2. (D) 2 (E) 1 Solution: The energy of simple harmonic motion is Ek Ep E = kA = + 2 2 1
When the oscillation passes through the equilibrium position, its energy is E=Ek=mv So we have A=v According to the problem, A'=2m/m=2A 4. A particle on a spring executes simple harmonic motion. When the particle is found at x=xma/2 the speed of the particle (A)v=vmax (B)=√3m/2()v=√2=12(D)2==2 Solution The energy of simple harmonic motion is E=kA=Ek+e When the osillation is at the equilibrium position, its energy is E=E=mane When the oscillation is at the maximum displacement, its energy is E=E.=ky2 So we have E=-k 42 According to the problem 1 E max 5. A particle on a spring executes simple harmonic motion. If the total energy of the particle is doubled then the magnitude of the maximum acceleration of the particle will increase by a factor of D (A)4 (B) (E)I(it remains unchanged) Solution The acceleration of the particle is a=-Ao cos(ot+o) Its maximum acceleration is a= do
When the oscillation passes through the equilibrium position, its energy is 2 2 1 E E mv = k = . So we have k m A = v . According to the problem, A k m A′ = 2v = 2 4. A particle on a spring executes simple harmonic motion. When the particle is found at x=xmax/2 the speed of the particle is ( B ) (A) max v v x = (B) 3 / 2 max v v x = (C) 2 / 2 max v v x = (D) vx = vmax / 2 Solution: The energy of simple harmonic motion is Ek Ep E = kA = + 2 2 1 When the oscillation is at the equilibrium position, its energy is 2 max 2 1 E E mv = k = When the oscillation is at the maximum displacement, its energy is 2 max 2 1 E E kx = p = So we have 2 max 2 max 2 2 1 2 1 2 1 E = kA = mv = kx According to the problem, max 2 max 2 2 max 2 max 2 2 3 2 1 4 1 2 1 2 1 ) 2 ( 2 1 2 1 mv mv mv v v x E = mv + k ⇒ = + ⋅ ⇒ = 5. A particle on a spring executes simple harmonic motion. If the total energy of the particle is doubled then the magnitude of the maximum acceleration of the particle will increase by a factor of ( D ) (A) 4. (B) 8 . (C) 2. (D) 2 (E) 1 (it remains unchanged). Solution: The acceleration of the particle is cos( ) 2 a = −Aω ωt + φ , Its maximum acceleration is 2 amax = Aω
The energy of simple harmonic motion is E==kA2 According to the problem E"=2E=2·kf2=kr2→A=√2A Then the maximum acceleration is a=402=v2A02 I. Filling the blanks 1. Two springs with spring constants k, and k2 are Frictionless connected as shown in Figure I with a mass m attached on a frictionless surface. We pull on m with a (00000 ook Surface force F and hold m at rest. the mass m moves a distance x and point a moves a distance x'from their equilibrium positions. By examining the forces on point A, you can get the relation of x and xis xk.+kx. If the system is modeled with a single spring stretching m a distance x, by examining the forces on m, the effective spring constant of the single spring is k, k2 k1+k2 According to the problem, we have F=F =F f =kx k, x'=k,(x-x) (x-x F.=F,=F→kx=kx→k1 2. A 500 g mass is undergoing simple harmonic oscillation that is described by the following equation for its position x(o) from equilibrium x((=(0.50m)cos(6.0 rad/s)t+2-rad The amplitude A of the oscillation is_0.5m_. The angular frequency o of the oscillation is 6.0 rad/s. The frequency v of the oscillation is_3 Hz The period Tof the oscillation is_ 1/3s
The energy of simple harmonic motion is 2 2 1 E = kA According to the problem E E kA kA A 2A 2 1 2 1 2 2 2 2 ′ = = ⋅ = ′ ⇒ ′ = Then the maximum acceleration is 2 2 max a′ = A′ω = 2Aω II. Filling the Blanks 1. Two springs with spring constants k1 and k2 are connected as shown in Figure 1 with a mass m attached on a frictionless surface. We pull on m with a force Fwe r and hold m at rest. The mass m moves a distance x and point A moves a distance x′ from their equilibrium positions. By examining the forces on point A, you can get the relation of x and x′ is x k k k x 1 2 2 + ′ = . If the system is modeled with a single spring stretching m a distance x, by examining the forces on m, the effective spring constant of the single spring is 1 2 1 2 k k k k + . Solution: According to the problem, we have x k k k k x k x x x F k x x F k x F F F k k k k we 1 2 2 1 2 2 1 ( ) ( ) 2 1 1 2 + ⇒ ′ = − ′ ⇒ ′ = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − ′ = ′ = = 1 2 1 2 1 2 2 1 1 ' 1 2 k k k k x kx k k k k F F F k x kx k k k we + = ⇒ = + = = ⇒ = ⇒ ⋅ 2. A 500 g mass is undergoing simple harmonic oscillation that is described by the following equation for its position x(t) from equilibrium: rad] 6.0 ( ) (0.50m)cos[(6.0π rad/s) π x t = t + The amplitude A of the oscillation is 0.5m . The angular frequency ω of the oscillation is 6.0π rad/s . The frequency ν of the oscillation is 3 Hz . The period T of the oscillation is 1/3 s . Fig.1 Frictionless Surface m k1 A k2 i ˆ
The spring constant k is 17747N/m. The position of oscillator when t=0 S is 0.43 m_. The time(t>0 s)the oscillator first at maximum distance from equilibrium is_0.44 S_. The maximum speed of the oscillator is_ 9.42 m/s_. The magnitude of the maximum acceleration of the oscillation is 17747 m/s2 Solution From the equation x(o)=(0.50m)cos[(. 0T rad/s)t+r rad 6.0 (a)A=0.5m (b)o=67 rad/s (c)v=2_6z 3HZ 2x2丌 6丌3 →k=a2m=(6x)2×05=182=17747N/n (When (=0 x=0.5cos =0433m g)=105co(60×/0→c06m+a)=1=67+x=kz k丌-2)k=0,±1,±2 Ifk=1t=-丌=0.44s (h)x=0.5c0s(6丌1+-) V=,=-3rsin(6t+-) v=3丌=942m/s )a==-18xc06+6
The spring constant k is 177.47 N/m . The position of oscillator when t = 0 s is 0.43 m . The time (t > 0 s ) the oscillator first at maximum distance from equilibrium is 0.44 s . The maximum speed of the oscillator is 9.42 m/s . The magnitude of the maximum acceleration of the oscillation is 177.47 m/s2 . Solution: From the equation ] 6.0 rad ( ) (0.50m) cos[(6.0π rad/s) π x t = t + , we get (a) A = 0.5m (b) ω = 6π rad/s (c) 3 Hz 2 6 2 = = = π π π ω ν (d) s 3 1 6 2 2 = = = π π ω π T (e) (6 ) 0.5 18 177.47 N/m 2 2 2 ω = ⇒ k = ω m = π × = π = m k (f) When t = 0 0.433 m 4 3 6 = 0.5cos = = π x g) π π π π π π x = πt + = ⇒ t + = ± ⇒ t + = k 6 ) 1 6 6 ) 0.5 cos(6 6 0.5cos(6 k t s t k k 0.44 36 5 If 1 ) 0, 1, 2, 6 ( 6 1 = = = ⇒ = − = ± ± π π π π K (h) ) 6 0.5cos(6 π x = π t + ) 6 3 sin(6 d d π = = − π π t + t x v 3 9.42 m/s vmax = π = (i) ) 6 18 cos(6 d d 2 π = = − π π + t v a
=18x2=17747m/s2 3. An oscillator consists of a block attached to a spring(k=456N/m). At some time L, the position (measured from the equilibrium location), velocity, and acceleration of the block are x=0.112m v-13.6m/s, a,-123m/s. The frequency of the oscillation is_ 5.28Hz, the mass of the block is 0. 42kg, the amplitude of oscillation is-0. 43 m- Solution The general equation about the oscillation is x= A cos(@t +o) v=-A@sin(at+o)(2) a=-A@ coS(@t+o)(3 Substitute x, v a to the eugtions, solving(1)and(3), we have @=33.14(N/kg m) 33.14 So the frequency of the oscillation is f-2T 2x3 14 S28(Hz) The mass of the block is m=k= 456=0.42 kg Solving(2)and (3), we have the amplitude of oscillation is 13.6 A12V30+01122=043m 4. A 10.0 kg mass attached to a spring is dragged at constant speed up rough surface (us=0.30, uk=0.25)inclined at 10%to the horizontal as in Figure 2. The spring is stretched 8.0 cm. The 0000 spring constant of the spring is_514.32 N/m_. If the spring drags the mass at constant speed down the slope, the stretch of the spring is 0.014m Solution The second law force diagrams of the mass are shown in figure (a)Up the slop Apply Newton's second law of motion, we have F =0 f=uN F=k△x
2 2 amax =18π =177.47 m/s 3. An oscillator consists of a block attached to a spring (k=456N/m). At some time t, the position (measured from the equilibrium location), velocity, and acceleration of the block are x = 0.112m, vx=-13.6m/s, ax=-123m/s2 . The frequency of the oscillation is 5.28Hz , the mass of the block is 0.42kg , the amplitude of oscillation is 0.43 m . Solution: The general equation about the oscillation is ⎪ ⎩ ⎪ ⎨ ⎧ = − + = − + = + cos( ) (3) sin( ) (2) cos( ) (1) 2 ω ω φ ω ω φ ω φ a A t v A t x A t Substitute x, v, a to the euqtions, solving (1) and (3), we have ω = 33.14(N/kg ⋅ m) So the frequency of the oscillation is 5.28(Hz) 2 3.14 33.14 2 = × = = π ω f The mass of the block is 0.42 kg 33.14 456 2 2 = = = ω k m Solving (2) and (3), we have the amplitude of oscillation is 0.112 0.43m 33.14 13.6 2 2 2 2 2 2 = + x = + = v A ω 4. A 10.0 kg mass attached to a spring is dragged at constant speed up rough surface (µs = 0.30, µk = 0.25) inclined at 10° to the horizontal as in Figure 2. The spring is stretched 8.0 cm. The spring constant of the spring is 514.32 N/m . If the spring drags the mass at constant speed down the slope, the stretch of the spring is 0.014m . Solution: The second law force diagrams of the mass are shown in figure. (a) Up the slop Apply Newton’s second law of motion, we have ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = ∆ = − = − − = F k x f N N mg F f mg µk cos10 0 sin10 0 o o F r Fig.2 10° f v F r mg v N v
4r (u4, cos10+sin 10)=514.32(N/m) b)Down the slop Apply Newton s second law of motion, we have F-f+ mg sin10°=0 N- mg cosl0°=0 f=uN F=k△x △x="(cosl0°-sin10°)=0.014(m) II. Give the solutions of the following problems 1. A 1.50 kg mass on a horizontal frictionless surface is attached to a horizontal spring with spring constant k=200 N/m. The mass is in equilibrium at x=0 m. The mass is released when t=0 s at coordinate x=0. 100 m with a velocity(2.00 m/s)i (a) Find the constants A, o and o in x(()=Acos(ot +o) b) Find the period Tof the oscillation. (c) Determine the maximum speed of the oscillation and the magnitude of the maximum acceleration (d) Plot x(o during the time interval 1=0 s and /=2T. Solution (a)From the problem, 0=1 =133. 3 N/kg.m. so substitute A Vm v1.5 0.100 m and 1=2.00 m/s to x(o=Acos(at +o), we have vo =2=-Aasin=-133.3Asing 0.12+( =0.(m) Solving the equations, we have 133.3 丌 e=am-13x33 Thinking of the direction of the velocity of the mass at FO s, we get
( cos10 + sin10 ) = 514.32 (N/m) ∆ ⇒ = o o k x mg k µ (b) Down the slop Apply Newton’s second law of motion, we have ( cos10 sin10 ) 0.014 (m) cos10 0 sin10 0 ⇒ ∆ = − = ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = ∆ ′ = − = − + = o o o o k k k mg x F k x f N N mg F f mg µ µ III. Give the Solutions of the Following Problems 1. A 1.50 kg mass on a horizontal frictionless surface is attached to a horizontal spring with spring constant k = 200 N/m. The mass is in equilibrium at x = 0 m. The mass is released when t = 0 s at coordinate x = 0.100 m with a velocity (2.00 m/s)i ˆ. (a) Find the constants A, ω and φ in x(t) = Acos(ωt + φ) . (b) Find the period T of the oscillation. (c) Determine the maximum speed of the oscillation and the magnitude of the maximum acceleration. (d) Plot x(t) during the time interval t = 0 s and t = 2T. Solution: (a) From the problem, 133.3 N/kg m 1.5 200 = = = ⋅ m k ω , so substitute A, ω, t= 0 s, ,and x = 0.100 m and v=2.00 m/s to x(t) = Acos(ωt + φ) , we have ⎩ ⎨ ⎧ = = − = − = = ω φ φ φ 2 sin 133.3 sin 0.1 cos 0 0 v A A x A Solving the equations, we have ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = − = − = + = 3 3 5 ) 133.3 arctan( ) 0.1(m) 133.3 2 0.1 ( 0 0 2 2 π φ π or x v A Thinking of the direction of the velocity of the mass at t= 0 s, we get 3 π φ = − . N v f v F r mg v
2丌 (b) The period Tof the oscillation is T 133.3 (c) The maximum speed of the oscillation is vmax =Ao=0.1x1333=1333 m/s The magnitude of the maximum acceleration is amax=Ao=0.1x1333=1.78 x10 m/s 2. A 5 13kg object moves on a frictionless surface under the influence of a spring with force constant88N/cm. The object is displaced 53. 5cm and given an initial velocity of 11. 2m/s back toward the equilibrium position. Find(a) the frequency of the motion, ( b)the initial potential energy of the system,(c)the initial kinetic energy, and(d) the amplitude of the motion Solution (a)The frequency of the motion is I 2.2l(Hz) 2丌2xVm (b) The initial potential energy of the system is PE=kx=-x988x05352=141.4(J) 2 (c) The initial kinetic energy of the system is KE=-mmv-=x5.13x11 2=28. 730) (D)As the total kinetic energy is E= PE+KE=kA, Thus the amplitude of the motion A=2PE+KE)2×(4144287 059(m) 3. Electrons in an oscilloscope are deflected by two mutually perpendicular electric forces in such a way that at any time t the displacement is given by x=Acos ot and y= Acos(at+o) Describe the path of the electrons and determine its equation when(a),=0,(b)8,=30 and(c),=90° (a)When =0, x=Acos ot, y= Acos t, Thus y=x, it is a straight line (b)When =30
(b) The period T of the oscillation is 0.54(s) 133.3 2 2 = = = π ω π T (c) The maximum speed of the oscillation is 0.1 133.3 13.33 m/s vmax = Aω = × = The magnitude of the maximum acceleration is 2 2 3 2 amax = Aω = 0.1×133.3 = 1.78×10 m/s 2. A 5.13kg object moves on a frictionless surface under the influence of a spring with force constant 9.88N/cm. The object is displaced 53.5cm and given an initial velocity of 11.2m/s back toward the equilibrium position. Find (a) the frequency of the motion, (b) the initial potential energy of the system, (c) the initial kinetic energy, and (d) the amplitude of the motion. Solution: (a) The frequency of the motion is 2.21(Hz) 2 1 2 = = = m k π π ω ν (b) The initial potential energy of the system is 988 0.535 141.4(J) 2 1 2 1 2 2 PE = kx = × × = (c) The initial kinetic energy of the system is 5.13 11.2 28.73(J) 2 1 2 1 2 2 KE = mv = × × = (D) As the total kinetic energy is 2 2 1 E = PE + KE = kA , Thus the amplitude of the motion is 0.59(m) 988 2( ) 2 (141.4 28.73) = × + = + = k PE KE A 3. Electrons in an oscilloscope are deflected by two mutually perpendicular electric forces in such a way that at any time t the displacement is given by x = Acosωt and cos( ) y y = A ωt +φ . Describe the path of the electrons and determine its equation when (a) o = 0 φ y , (b) o = 30 φ y , and (c) o = 90 φ y . Solution: (a) When o = 0 φ y , x = Acosωt , y = Acosωt , Thus y=x, it is a straight line. (b) When o = 30 φ y , ⎪ ⎩ ⎪ ⎨ ⎧ = + = ) 6 cos( cos π ω ω y A t x A t
Solving it, we get x2+y2-v3xy==A2, it is a ellipse x= Acos ot y=Acos(ou Solving it, we get =I it is a circular 4. Two masses are undergoing simple harmonic oscillation according to the equations 1(1)=(0.150m)cos(0.250rads)t+/6rad] x2(l)=(0.200m)cos(0.350rads)t+/4rad a)Find the period of each oscillation b)On the same graph, plot xI(o) and x2(0) versus t during t=0 s to 60 s and determine the first six instants at which they have the same position Is the interval between these instants the same? (c)By changing one of the amplitudes, see if the instants when the oscillators have the same position depend on the amplitude 2丌2丌 (a)T 0.25 2 40丌 72 1794 0.357 7=0.36sx=x2=0.12m 9.0ls -0.14m 0.004m 3943s 49.40s0.14m 57.78s 0.11m The interval is not same (c)The instants when the oscillation has the same position depend on the amplitudes 5. A spring with spring constant k is attached to a mass m that is confined to move along a frictionless rail oriented perpendicular to the lo kio axis of the spring as indicated in Figure 3. The spring is initially unstretched and of length lo when the mass is at the position x =0 m in the indicated coordinate system. Show that when the mass is released from the point x along the rail, the oscillations occur but their
Solving it, we get 2 2 2 4 1 x + y − 3xy = A , it is a ellipse. (c) When o = 90 φ y , ⎪ ⎩ ⎪ ⎨ ⎧ = + = ) 2 cos( cos π ω ω y A t x A t Solving it, we get 1 2 2 2 2 + = A y A x , it is a circular. 4. Two masses are undergoing simple harmonic oscillation according to the equations x1(t)=(0.150m) cos [(0.250 rad/s) t + π/6 rad] x2(t)=(0.200m) cos [(0.350 rad/s) t + π/4 rad] (a) Find the period of each oscillation. (b) On the same graph, plot x1(t) and x2(t) versus t during t = 0 s to 60 s and determine the first six instants at which they have the same position. Is the interval between these instants the same? (c) By changing one of the amplitudes, see if the instants when the oscillators have the same position depend on the amplitudes. Solution: (a) 8 25.12 s 0.25 2 2 1 1 = = = π = π ω π T 17.94 s 7 40 0.35 2 2 2 2 = = = = π π ω π T (b) T=0.36 s x1= x2=0.12 m 9.01 s -0.14 m 29.22 s 0.004 m 39.43 s 0.09 m 49.40 s 0.14 m 57.78 s -0.11 m The interval is not same. (c) The instants when the oscillation has the same position depend on the amplitudes. 5. A spring with spring constant k is attached to a mass m that is confined to move along a frictionless rail oriented perpendicular to the axis of the spring as indicated in Figure 3. The spring is initially unstretched and of length l0 when the mass is at the position x = 0 m in the indicated coordinate system. Show that when the mass is released from the point x along the rail, the oscillations occur but their Fig.3 l0 k x m θ O
oscillations are not simple harmonic oscillations Solutio The mass is pulled out a distance x along the rail, the new total length of the spring is So the x-component of the force that the spring exerted on the mass is F=FcosB=-k[( +x)-lo]cos 8 From the diagram, cos so the x-component of the force +l0 F=-k(x )=-kx(1 )=-kx(1 For x<<lo=()<<l, the x-component of the force F≈-k1-1()2+1 +1=1 hen F s-k(1-11 x2 Thus when the mass is released from the point x along the rail, the oscillations occur but their oscillations are not simple harmonic oscillations
oscillations are not simple harmonic oscillations. Solution: The mass is pulled out a distance x along the rail, the new total length of the spring is 2 2 0 l = l + x So the x-component of the force that the spring exerted on the mass is cosθ [( ) 0 ]cosθ 2 1 2 2 0 F F k l x l x = = − + − From the diagram , 2 0 2 cos x l x + θ = , so the x-component of the force is ) ( ) 1 1 ( ) (1 2 0 2 0 2 0 + = − − + = − − l x kx x l l x F k x x ) ( ) 1 ( ) 1 (1 2 0 2 0 + + = − − l x l x kx For ( ) 1 2 0 << 0 ⇒ << l x x l , the x-component of the force is [1 ( ) 1] 2 0 ≈ − − + l x F kx x Since + = − 2 −L 0 2 0 ( ) 2 1 ( ) 1 1 l x l x Then 3 2 0 2 0 2 2 ) 2 1 (1 1 x l k l x F kx x ≈ − − + ⋅ = − Thus when the mass is released from the point x along the rail, the oscillations occur but their oscillations are not simple harmonic oscillations