Question

A simple harmonic motion having an amplitude A and time period T is represented by the equation:
$y=5sin\pi (t+4)$ m. Then the values of A (in m) and T (in sec) are:

• A. $A=5,T=2$
• B. $A=1,T=1$
• C. $A=5,T=1$
• D. $A=10,T=2$

Answer 1

The correct option is A $A=5,T=2$

Comparing the equation of SHM with the standard equation,

The amplitude of the oscillation will be $a=5m$.

The angular frequency of the oscillation will be $\omega =\pi$

So, time period will be:

$T=\frac{2\pi }{\omega }$

$T-\frac{2\pi }{\pi }$

$=2sec$