Question

An object undergoing simple harmonic motion takes 0.40 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.

Answer 1

Answer:

a)  $P=0.80$

b)  $1.25Hz$

c)  $A=25cm$

Explanation:

From the question we are told that:

Travel Time $T=0.40s$

Distance $d=50cm$

a)

Period

Time taken to complete one oscillation

Therefore

$P=2*T\\\\P=2*0.40$

$P=0.80$

b)

Frequency is

$F=\frac{1}{T}\\\\F=\frac{1}{0.80}$

$1.25Hz$

c)

Amplitude:the distance between the mean and extreme position

$A=\frac{50}{2}$

$A=25cm$