Question

A sinusoidal wave (transverse) is produced on a string with frequency $f$ and amplitude $A$. The ratio of velocity amplitude to acceleration amplitude is

• A. $\frac{\pi }{f}$
• B. $\frac{2\pi }{f}$
• C. $\frac{1}{\pi f}$
• D. $\frac{-1}{2\pi f}$

Answer 1

Answer: (D)

$\frac{-1}{2\pi f}$

$y=A\sin (\omega t+kx)$
$\omega =2\pi f$
$y=A\sin (2\pi ft+kx)$
$V=\frac{dy}{dl}=A\times 2\pi f\cos (2\pi ft+kx)$
$a=\frac{dv}{dt}=-A\left(2\pi t{)}^2\sin \right(2\pi ft+kx)$
$\therefore \frac{V}{a}=\frac{-1}{2\pi f}$