Question

Vibration in a string of length $60\text{cm}$ fixed at both ends is represented by the equation, $y=4\sin \left(\frac{\pi x}{15}\right)\cos \left(96\pi t\right)$, where $x$ and $y$ are in $\text{centimeter}$ and $t$ in $\text{seconds}$. The number of loops formed in vibrating string will be

• A. $4$
• B. $3$
• C. $6$
• D. $5$

Answer 1

The correct option is A $4$

The equation of stationary wave in string is,
$y=4\sin \left(\frac{\pi x}{15}\right)\cos \left(96\pi t\right)$
Comparing above equation with
$y=2A\sin kx\cos \omega t$
We get,
$k=\frac{\pi }{15}{\text{cm}}^{-1}$
$\Rightarrow \lambda =\frac{2\pi }{k}$
$\Rightarrow \lambda =30\text{cm}$
For one loop of stationary wave, $l=\frac{\lambda }{2}=15\text{cm}$
Since string is fixed at both ends, they will act as nodes. Let $\left(n\right)$ loops of stationary wave will be formed between both nodes.
$\Rightarrow L=nl$
where $L\to$ length of the string.
$\Rightarrow n=\frac{L}{l}$
$\therefore n=\frac{60}{15}=4$