Question

The equation of a particle executing SHM is $y=20\sin (2\pi t-\frac{\pi }{3})$.
The amplitude $A$, time period $T$ and angular frequency $\omega$ are
[All quantities are in SI units]

• A. $A=10\text{m},T=1\text{sec},\omega =2\pi \text{rad/sec}$
• B. $A=20\text{m},T=2\text{sec},\omega =2\pi \text{rad/sec}$
• C. $A=40\text{m},T=1\text{sec},\omega =2\pi \text{rad/sec}$
• D. $A=20\text{m},T=1\text{sec},\omega =2\pi \text{rad/sec}$

Answer 1

The correct option is D $A=20\text{m},T=1\text{sec},\omega =2\pi \text{rad/sec}$

Given,
$y=20\sin (2\pi t-\frac{\pi }{3})$
Comparing it with the general equation of SHM,
$y=A\sin (\omega t+\varphi )$
$A=20\text{m},\omega =2\pi \text{rad/sec},T=\frac{2\pi }{\omega }=\frac{2\pi }{2\pi }\Rightarrow T=1\text{sec}$