Question

A particle of mass $2kg$ is performing SHM, given by equation $x=10\sin 2\pi t$ ($x$ is in $m$ and $t$ in $s$). The work done on the particle in time $0.25s$ to $0.75s$, will be

• A. $\begin{array}{l}200{\pi }^{2}J\end{array}$
• B. $\begin{array}{l}100{\pi }^{2}J\end{array}$
• C. $\begin{array}{l}50{\pi }^{2}J\end{array}$
• D. Zero

Answer 1

The correct option is D

Zero

Step 1: Given Data

Displacement in SHM,

$x=10\sin 2\pi t$

Mass of the particle $m=2kg$

Time $t_1=0.25s$

Time $t_2=0.75s$

Let the frequency of SHM be $\omega$.

Step 2: Formula Used

Acceleration of a particle in SHM is given as,

$a=-{\omega }^{2}x$

Force on a particle is given as,

$F=ma$

Work done by the particle is given as,

$W={\int }_{x_1}^{x_2}F.dx$

Step 3: Calculate the Force

From the given displacement in SHM $x=10\sin 2\pi t$ we get,

$\omega =2\mathrm{\pi }$

Therefore, acceleration

$$\begin{gathered} a=-{\omega }^{2}x \\ =-{\left(2\mathrm{\pi }\right)}^{2}x \\ =-4{\mathrm{\pi }}^2\mathrm{x} \end{gathered}$$

Therefore, force,

$$\begin{gathered} F=ma \\ =2\times \left(-4{\mathrm{\pi }}^2\mathrm{x}\right) \\ =-8{\mathrm{\pi }}^2\mathrm{x} \end{gathered}$$

Step 4: Calculate the Limits

For time $t_1=0.25s$,

$$\begin{gathered} x_1=10\sin \left(2\mathrm{\pi }\times 0.25\right) \\ =10\sin \frac{\mathrm{\pi }}{2}=10m \end{gathered}$$

Similarly, for time $t_2=0.75s$

$$\begin{gathered} x_2=10\sin \left(2\mathrm{\pi }\times 0.75\right) \\ =10\sin \frac{3\mathrm{\pi }}{2}=-10m \end{gathered}$$

Step 5: Calculate the Work Done

$$\begin{gathered} W={\int }_{x_1}^{x_2}F.dx \\ ={\int }_{10}^{-10}-8{\mathrm{\pi }}^2\mathrm{x}.dx \\ =8{\mathrm{\pi }}^2{\int }_{-10}^{10}\mathrm{x}.dx \\ =8{\mathrm{\pi }}^2{\overline{)\frac{{\mathrm{x}}^2}{2}}}_{-10}^{10} \\ =8{\mathrm{\pi }}^2\left[\frac{{\left(10\right)}^2-{\left(-10\right)}^2}{2}\right] \\ =8{\mathrm{\pi }}^2\left[\frac{100-100}{2}\right]=0 \end{gathered}$$

Hence, the correct answer is option (D).