Question

The displacement - time relation for a particle can be expressed as $y=0.5\left[{\cos }^2\right(n\pi t)-{\sin }^2(n\pi t\left)\right]$. This relation shows that

• A. the particle is executing SHM with amplitude $0.5\text{m}$
• B. the particle is executing SHM with a frequency $n$ times that of a second's pendulum
• C. the particle is executing SHM and the velocity at its mean position is $\left(n\pi \right)\text{m/s}$
  
• D. the particle is not executing SHM at all

Answer 1

Answer: (A), (C)

the particle is executing SHM and the velocity at its mean position is $\left(n\pi \right)\text{m/s}$


Given:
The displacement - time relation is,

$y=0.5\left[{\cos }^2\right(n\pi t)-{\sin }^2(n\pi t\left)\right]$

$(\because \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta )$

$\therefore y=0.5\cos \left(2n\pi t\right).....\left(1\right)$

after differentiating w.r.t. $t$, we get

$\frac{dy}{dt}=-0.5\times 2n\pi \times \sin \left(2n\pi t\right)...\left(2\right)$

again differentiating

$\frac{d^{2}y}{d{t}^2}=-\left(0.5\right)\left(2n\pi {)}^2\cos \right(2n\pi t)$

$\Rightarrow \frac{d^{2}y}{d{t}^2}=-4n^2{\pi }^{2}y\text{(from eq. (1))}$

Since, $n$ and $\pi$ are constant

$\therefore \frac{d^{2}y}{d{t}^2}\propto -y$

i.e., particle is executing SHM with amplitude $0.5\text{m}$.

That means choice $\left(a\right)$ is correct.

As standard equation of SHM is

$\frac{d^{2}y}{d{t}^2}=-{\omega }^{2}y$

Hence, $\omega =2n\pi$

For a second's pendulum, $T=2\text{s}$

So, ${\omega }^{\prime }=\frac{2\pi }{T}=\pi$

$\therefore \omega =2n\times {\omega }^{\prime }$

Means option $\left(b\right)$ is not correct.

Now, from equation (2)

${\left(\frac{dy}{dt}\right)}_{max}=n\pi$

i.e., choice $\left(c\right)$ is also correct.