Question

A force $F=-10x+2$ acts on a particle of mass $0.1$kg, where '$k$' is in m and $F$ in newton.If it is released from rest at $x=-2$m,find:(a) amplitude; (b) time period; (c) equation of motion.

• A. (a)$\frac{11}{5}m$ (b)$\frac{\pi }{5}sec$ (c) $x=0.2-\frac{11}{5}cos\omega t$
• B. (a)$\frac{11}{3}m$ (b)$\frac{\pi }{5}sec$ (c) $x=0.2-\frac{11}{5}cos\omega t$
• C. (a)$\frac{11}{5}m$ (b)$\frac{\pi }{3}sec$ (c) $x=0.2-\frac{11}{5}cos\omega t$
• D. (a)$\frac{11}{5}m$ (b)$\frac{\pi }{5}sec$ (c) $x=0.2-\frac{11}{3}cos\omega t$

Answer 1

The correct option is A (a)$\frac{11}{5}m$ (b)$\frac{\pi }{5}sec$ (c) $x=0.2-\frac{11}{5}cos\omega t$

Force, $F=-10x+2$

$F=-10(x-\frac{1}{5})$

Let $y=x-\frac{1}{5}$, $F=-10y$

Now $F=-kx$

So, $k=10N/m$

a) Amplitude = Maximum displacement $=|y|=|x-\frac{1}{5}|=|(-2)-\frac{1}{5}|=\frac{11}{5}$

b) Now, $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{10}{0.1}}=10rad/s$

Time period, $T=\frac{2\pi }{\omega }=\frac{2\pi }{10}=\frac{\pi }{5}sec$

c) Finally, equation of motion will be given by:

$y=Acos(\omega t+\varphi )$

Now at $t=0$ when displacement is maximum.

$-\frac{11}{5}=\frac{11}{5}cos\varphi$

So, $\varphi =co{s}^{-1}(-1)=\pi$

Therefore, $y=Acos(\omega t+\pi )=-Acos\left(\omega t\right)$

Substitute for y in terms of x

$x-\frac{1}{5}=-\frac{11}{5}cos\left(\omega t\right)$

$x=\frac{1}{5}-\frac{11}{5}cos\left(\omega t\right)$