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