An object undergoing simple harmonic motion takes 0.25 sto travel from one point of zero velocity to the next such point. The distance between those points is 36 cm.

(a) Calculate the period of the motion.

(b) Calculate the frequency of the motion.

(c) Calculate the amplitude of the motion.

i) Time to travel from zero-point velocity to next such point,$t\mathit{=}0.25\mathrm{s}$

ii) Distance between point having zero velocity to the next point,$D=36\mathrm{cm}$

We are given the time between points having zero velocity. So, the time of motion is half of that time. Once we find the time of motion, we can use it to find the frequency. Also, as the distance between points having zero velocity is given, the amplitude is half of that distance.

Formula:

The frequency related to the time period of a body,

$f=\frac{1}{T}$ …(i)

The period of motion is the time required to complete one cycle. The time between points having zero velocity is already given.

So,thetime period of motion is twicethe timebetween points having zero velocity.

Hence, the time period is given by:

$$\begin{gathered} T=0.25\sec \times 2\left(\because \mathrm{T}=2\mathrm{t}\right) \\ =0.5\sec \end{gathered}$$

Hence, the time period is 0.5 sec.

In the displacement versus time graph, points A and B are points having zero velocity.

The frequency of motion is calculated using equation (i) as follows:

f =1/(0.5 sec)

=2 Hz

Hence, the value of frequency is 2 Hz

The distance between points having zero velocity is already given, and the amplitude is half of that distance.

So, amplitude is A

$$\begin{gathered} A=\frac{d}{2} \\ =\frac{36\mathrm{cm}}{2} \\ =18\mathrm{cm} \end{gathered}$$

Hence, the amplitude of motion is 18 cm.