Question

The equation of a particle executing SHM is $x=\left(5m\right)sin\left[\right(\pi s^{-1})t+\frac{\pi }{3}]$. What is its amplitude, time period, maximum speed and velocity. At t = 1s?

• A. $5m,2s,5m{s}^{-1},\frac{-5\pi }{2}m{s}^{-1}$
• B. $5m,2s,5\pi m{s}^{-1},\frac{-5\pi }{2}m{s}^{-1}$
• C. $5m,\pi s,5m{s}^{-1},\frac{-5\pi }{2}m{s}^{-1}$
• D. $5m,2s,5\pi m{s}^{-1},\frac{+5\pi }{2}m{s}^{-1}$

Answer 1

The correct option is B

$5m,2s,5\pi m{s}^{-1},\frac{-5\pi }{2}m{s}^{-1}$

A typical equation representing SHM looks like $x=Asin(\omega t+{\psi }_0)$

Where x represents displacement

$\omega$ represents angular frequency

${\psi }_0$ represents initial phase

A represents amplitude

In the question we have $x=\left(5m\right)sin\left[\right(\pi s^{-1})t+\frac{\pi }{3}]$

So let's compare both! In a simple glance one can see

$A=5m,\omega =\pi s^{-1}and{\psi }_0=\frac{\pi }{3}$

Time period $\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2s$

Now what about velocities?

Let's differentiate the equation $\frac{dx}{dt}=\left(5m\right)\pi cos\left\{\right(\pi s^{-1})t+\frac{\pi }{3}\}$

$v=\left(5\pi \right)mcos\left\{\right(\pi s^{-1})t+\frac{\pi }{3}\}$

We know that $\frac{dx}{dt}$ means velocity. To find the maximum velocity we can argue that cos has maximum value of 1 so $V_{max}=5\pi m{s}^{-1}$. To

find velocity at t = 1s, let's substitute for t in velocity equation we get $V_{1s}=\frac{-5\pi m{s}^{-1}}{2}$.