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= doWhen 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