Q. A simple harmonic wave having an amplitude $A$ and time period $T$ is represented by the equation $y = 5\ sin\ \pi(t + 4)m$. Then the value of $A$ in (metre) and $T$ in (second) are
A) $A = 10,T = 2$
B) $A = 5,\ T = 1$
C) $A = 10,\ T = 1$
D) $A = 5,\ T = 2$
Explanation
(d) The equation is given
$y = 5\sin{\pi(t + 4)}$ …(i)
The standard equation is
$y = A\sin{(\omega t + \varnothing)}$ …(ii)
Now comparing Eqs. (i) and (ii), we get
$A = 5\ m,\ \ \omega t = \pi t\ \ \ \ \ or\ \ \ \omega = \pi$
So, $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2$ s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.