=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△x2 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