A particle attached to a spring with k = 51 N/m is undergoing simple harmonic mo

Question

A particle attached to a spring with k = 51 N/m is undergoing simple harmonic motion, and its position is described by the equation x = (5.2 m)cos(8.2t), with t measured in seconds. (a) What is the mass of the particle? kg (b) What is the period of the motion? s (c) What is the maximum speed of the particle? m/s (d) What is the maximum potential energy? J (e) What is the total energy? J

Explanation / Answer

A particle attached to a spring with k = 48 N/m is undergoing simple harmonic motion, and its position is described by the equation x = (5.5 m)cos(7.6t), with t measured in seconds. (a) What is the mass of the particle? (b) What is the period of the motion? (c) What is the maximum speed of the particle? (d) What is the maximum potential energy? (e) What is the total energy? We are given the spring constant k=48 N/m, the amplitude (maximum displacement) A=5.5m, the angular frequency w=7.6 rad/s, and the equation for the position (displacement) x=A cos(wt). (a) The angular frequency is related to the spring constant and the mass (m) of the particle by the equation w=sqrt(k/m) => m=k/w^2=(48 N/m)/(7.6/s)^2=0.831 kg (Note: 1 N=1 kg m/s^2) (b) The angular frequency is related to the period (T) by w=2 pi/T => T=2 pi/w=2*3.14/(7.6/s)=0.826 s (c) The instantaneous velocity (v) of an object is given by the time derivative of its position, i.e., v=dx/dt=d[A cos(wt)]/dt=A d[cos(wt)]/dt=A w [-sin(wt)]=-A w sin(wt) The maximum speed occurs when sin(wt)=-1, so that vmax=A w=5.5m*7.6/s=41.8m/s (d) The potential energy (U) stored in a spring is given by U=(k x^2)/2=k [(A cos(wt))^2]/2=k A^2 cos^2(wt)/2 The potential energy is a maximum when cos(wt)=+/-1, so Umax=(k A^2)/2=48 N/m*((5.5m)^2)/2=726 J (Note: 1 J=1 N m= 1 kg m^2/s^2) (e) The total energy (E) of the system is the sum of the potential energy (U) and the kinetic energy (K). Recall that K=(m v^2)/2 and m=k/w^2, so that K=[m (-A w sin(wt))^2]/2=(k/w^2)[A^2 w^2 sin^2(wt)]/2=k A^2 sin^2(wt)/2 In part (d) we found that U=k A^2 cos^2(wt)/2 Thus, E=K+U=k A^2 sin^2(wt)/2+k A^2 cos^2(wt)/2=[(k A^2)/2][sin^2(wt)+cos^2(wt)] Recall that sin^2 (B)+cos^2 (B)=1 So E=(k A^2)/2=Umax=726 J

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