Question

An object is undergoing SHM with period 0.960 s and amplitude 0.320 m. At t= 0, the object is at x = 0.320 and is instantaneously at rest.

A) Calculate the time it takes the object to go from x = 0.320 m, to x = 0.160 m.

B) Calculate the time it takes the object to go from x= 0.160 m, to x = 0 m.

Answer 1

Concepts and reason

The concepts used in this question are equation for simple harmonic motion, and properties of simple harmonic motion.

Firstly, write the expression of the displacement of object under simple harmonic motion and then substitute the values in it to find the time.

Fundamentals

The displacement of an object undergoing a simple harmonic motion is given by the following expression:

$x\left( t \right) = A\cos \left( {\omega t} \right)$

Here, $\omega$ is the angular frequency and A is the amplitude of the motion.

The expression for the angular frequency is as follows:

$\omega = \frac{{2\pi }}{T}$

Here, T is the time-period of oscillation.

(a)

The expression for the angular frequency is as follows:

$\omega = \frac{{2\pi }}{T}$

Substitute 0.960 s for T in the above expression.

$\begin{array}{c}\\\omega = \frac{{2\pi }}{{0.960{\rm{ s}}}}\\\\ = 6.545{\rm{ rad/s}}\\\end{array}$

The displacement of an object undergoing a simple harmonic motion is given by the following expression:

$x\left( t \right) = A\cos \left( {\omega t} \right)$

Rearrange the above expression for t.

$t = \frac{1}{\omega }{\cos ^{ - 1}}\left( {\frac{{x\left( t \right)}}{A}} \right)$

Substitute 0.320 m for A, 6.545 rad/s for $\omega$ , and 0.160 m for $x\left( t \right)$ in the above expression.

$\begin{array}{c}\\t = \frac{1}{{\left( {6.545{\rm{ rad/s}}} \right)}}{\cos ^{ - 1}}\left( {\frac{{0.160{\rm{ m}}}}{{0.320{\rm{ m}}}}} \right)\\\\ = \frac{1}{{\left( {6.545{\rm{ rad/s}}} \right)}}{\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\\\\ = \frac{1}{{\left( {6.545{\rm{ rad/s}}} \right)}}\left( {\frac{\pi }{3}} \right)\\\\ = 0.160{\rm{ s}}\\\end{array}$

(b)

The displacement of an object undergoing a simple harmonic motion is given by the following expression:

$x\left( t \right) = A\cos \left( {\omega t} \right)$

Rearrange the above expression for t.

$t = \frac{1}{\omega }{\cos ^{ - 1}}\left( {\frac{{x\left( t \right)}}{A}} \right)$

Substitute 0.320 m for A, 6.545 rad/s for $\omega$ , and 0.00 m for $x\left( t \right)$ in the above expression.

$\begin{array}{c}\\t = \frac{1}{{\left( {6.545{\rm{ rad/s}}} \right)}}{\cos ^{ - 1}}\left( {\frac{{0.000{\rm{ m}}}}{{0.320{\rm{ m}}}}} \right)\\\\ = \frac{1}{{\left( {6.545{\rm{ rad/s}}} \right)}}{\cos ^{ - 1}}\left( 0 \right)\\\\ = \frac{1}{{\left( {6.545{\rm{ rad/s}}} \right)}}\left( {\frac{\pi }{2}} \right)\\\\ = 0.240{\rm{ s}}\\\end{array}$

The time taken by the block to reach the mean position from the extreme position is 0.24 s.

The time taken by the object to reach 0.160 m from extreme position is 0.160 s. thus, the time taken by the object to reach the mean position from 0.160 m is as follows:

$\begin{array}{c}\\t = 0.240{\rm{ s}} - 0.160{\rm{ s}}\\\\ = 0.08{\rm{ s}}\\\end{array}$

Ans: Part a

The time taken by the object to reach 0.160 m from 0.320 m is 0.160 s.

Part b

The time taken by the object to reach 0.00 m from 0.160 m is 0.0800 s.