Wave Motion 2 Question 41
41. A string of mass per unit length $\mu$ is clamped at both ends such that one end of the string is at $x=0$ and the other is at $x=l$. When string vibrates in fundamental mode amplitude of the mid-point $O$ of the string is $a$, and tension in the string is $T$. Find the total oscillation energy stored in the string.
(2003, 4M)
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Answer:
Correct Answer: 41. $\frac{\pi^{2} a^{2} T}{4 l}$
Solution:
- $l=\frac{\lambda}{2}$ or $\lambda=2 l, k=\frac{2 \pi}{\lambda}=\frac{\pi}{l}$
The amplitude at a distance $x$ from
$x=0$ is given by $A=a \sin k x$
Total mechanical energy at $x$ of length $d x$ is
$$ \begin{aligned} d E & =\frac{1}{2}(d m) A^{2} \omega^{2} \\ & =\frac{1}{2}(\mu d x)(a \sin k x)^{2}(2 \pi f)^{2} \end{aligned} $$
or
$$ d E=2 \pi^{2} \mu f^{2} a^{2} \sin ^{2} k x d x $$
Here,
$$ f^{2}=\frac{v^{2}}{\lambda^{2}}=\frac{\left(\frac{T}{\mu}\right)}{\left(4 l^{2}\right)} \text { and } k=\frac{\pi}{l} $$
Substituting these values in Eq. (i) and integrating it from $x=0$ to $x=l$, we get total energy of string
$$ E=\frac{\pi^{2} a^{2} T}{4 l} $$
