Question

A particle executes SHM. along a straight line with mean position $x = 0$, period $20\,{\text{s}}$ and amplitude $5\,{\text{cm}}$. The shortest time taken by the particle to go from $x = 4\,{\text{cm}}$ to $x = - 3\,{\text{cm}}$ is:
A. $4\,{\text{s}}$
B. $7\,{\text{s}}$
C. $5\,{\text{s}}$
D. $6\,{\text{s}}$

Answer 1

Use the formula for the displacement of position of the particle. This formula gives the relation between the position, amplitude of particle, angular frequency of particle, time and phase difference. First determine the phase of the particle at position $x = 4\,{\text{cm}}$ and then determine the total time required for reach position $x = - 3\,{\text{cm}}$.

Formula used:
The displacement $x$ of a particle executing simple harmonic motion is given by
$x = A\sin \left( {\omega t + \phi } \right)$ …… (1)
Here, $A$ is amplitude of the particle, $\omega$ is angular frequency of the particle, $t$ is time and $\phi$ is phase difference.
The angular frequency $\omega$ of a particle is given by
$\omega = \dfrac{{2\pi }}{T}$ …… (2)
Here, $T$ is the time period of the particle.

Complete step by step answer:
We have given that the period of oscillation of the particle executing simple harmonic motion is $20\,{\text{s}}$ and the amplitude of motion is $5\,{\text{cm}}$.
$T = 20\,{\text{s}}$
$A = 5\,{\text{cm}}$

Let us first determine the angular frequency of this particle.
Substitute $20\,{\text{s}}$ for $T$ in equation (2).
$\omega = \dfrac{{2\pi }}{{20\,{\text{s}}}}$
$\Rightarrow \omega = \dfrac{\pi }{{10}}\,{\text{rad/s}}$
Hence, the angular frequency of the particle is $\dfrac{\pi }{{10}}\,{\text{rad/s}}$.

We have asked to determine the minimum time required for the particle to move from $x = 4\,{\text{cm}}$ to $x = - 3\,{\text{cm}}$.

Let us suppose that the particle starts oscillating from the position $x = 4\,{\text{cm}}$. Hence, the particle is at position$x = 4\,{\text{cm}}$ at time $t = 0\,{\text{s}}$.
Substitute $4\,{\text{cm}}$ for $x$, $5\,{\text{cm}}$ for $A$ and $0\,{\text{s}}$ for $t$ in equation (1).
$4\,{\text{cm}} = 5\,{\text{cm}}\sin \left( {\omega \left( {0\,{\text{s}}} \right) + \phi } \right)$
$\Rightarrow \sin \phi = \dfrac{4}{5}$
$\Rightarrow \phi = {\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right)$
This gives the phase of the particle.

Now let us assume that the particle reaches the position $x = - 3\,{\text{cm}}$ at time ${t_1}$.
Substitute $- 3\,{\text{cm}}$ for $x$, $5\,{\text{cm}}$ for $A$ and ${t_1}$ for $t$ in equation (1).
$- 3\,{\text{cm}} = 5\,{\text{cm}}\sin \left( {\omega {t_1} + \phi } \right)$
$\Rightarrow \sin \left( {\omega {t_1} + \phi } \right) = \dfrac{{ - 3}}{5}$

Substitute $\dfrac{\pi }{{10}}\,{\text{rad/s}}$ for $\omega$ and ${\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right)$ for $\phi$ in the above equation.
$\sin \left[ {\left( {\dfrac{\pi }{{10}}\,{\text{rad/s}}} \right){t_1} + {{\sin }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right] = \dfrac{{ - 3}}{5}$
$\Rightarrow \left[ {\left( {\dfrac{\pi }{{10}}\,{\text{rad/s}}} \right){t_1} + {{\sin }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right] = {\sin ^{ - 1}}\left( {\dfrac{{ - 3}}{5}} \right)$
$\Rightarrow \dfrac{\pi }{{10}}{t_1} = {\sin ^{ - 1}}\left( {\dfrac{{ - 3}}{5}} \right) - {\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right)$
$\Rightarrow {t_1} = 10\left[ {\dfrac{{{{\sin }^{ - 1}}\left( {\dfrac{{ - 3}}{5}} \right) - {{\sin }^{ - 1}}\left( {\dfrac{4}{5}} \right)}}{\pi }} \right]$
$\Rightarrow {t_1} = 10\left[ {\dfrac{{ - 0.643 - 0.927}}{{3.14}}} \right]$
$\Rightarrow {t_1} = - 5\,{\text{s}}$
$\therefore {t_1} = 5\,{\text{s}}$
Therefore, the shortest time required for the particle to travel is $5\,{\text{s}}$.

Hence, the correct option is C.

Note: From the above calculations, the time required for motion is found to be –5 seconds. But the time can never be negative. Hence, we have taken the positive value of the time as 5 second. We have taken the time zero at position 4 cm because we have supposed that the particle starts its motion from position 4 cm hence the time is zero at this position.