Question

please double check the answers of Question 1 C and Question 2 D i got the rest of the answers.

1.An object undergoing simple harmonic motion takes 0.32 s to travel from one point of zero velocity to the next such point. The distance between those points is 26 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.

2.x = (9.2 m) cos[(5πrad/s)t + π/4 rad]
gives the simple harmonic motion of a body. At t = 2.1 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?

Answer 1

1C. Extreme points have zero velocities. So the distance between the two extremes is 26cm. So the amplitude is the distance from the point of maximum velocity to the point of zero velocity. And it should have half the distance between two extremes points. So the amplitude is 13cm.

2D. As the equation for simple harmonic motion is given,i.e.   x = (9.2 m) cos[(5πrad/s)t + π/4 rad]

According to the given equation of motion the phase is π/4rad. And it same at any given time.