Wave Motion Ques 58
- The vibrations of a string of length $60 cm$ fixed at both ends are represented by the equation
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Answer:
Correct Answer: 58.(a) $2 \sqrt{3} cm \quad$ (b) $x=0,15 cm, 30 cm \ldots$ etc
(c) Zero
(d) $y _1=2 \sin \left(\frac{\pi x}{15}-96 \pi t\right)$ and $y _2=2 \sin \left(\frac{\pi x}{15}+96 \pi t\right)$
Solution:
Formula:
Vibrations of Strings (Standing Wave):
- (a) $y=4 \sin \frac{\pi x}{15} \cos (96 \pi t)=A _x \cos (96 \pi t)$
Here, $A _x=4 \sin \frac{\pi x}{15}$
at $x=5 cm, A _x=4 \sin \frac{5 \pi}{15}=4 \sin \frac{\pi}{3}=2 \sqrt{3} cm$
This is the amplitude or maximum displacement at
$$ x=5 cm \text {. } $$
(b) Nodes are located where $A _x=0$
$$ \text { or } \frac{\pi x}{15}=0, \pi, 2 \pi \ldots \ldots \text { or } x=0,15 cm, 30 cm etc . $$
(c) Velocity of particle,
$$ v _p=\left|\frac{\partial y}{\partial t}\right| _{x=\text { constant }}=-384 \pi \sin \left(\frac{\pi x}{15}\right) \sin (96 \pi t) $$
At $x=7.5 cm$ and $t=0.25 s$
$$ v _p=-384 \pi \sin \left(\frac{\pi}{2}\right) \sin (24 \pi)=0 $$
(d) Amplitude of components waves is $A=\frac{4}{2}=2 cm$
$$ \omega=96 \pi \text { and } k=\frac{\pi}{15} $$
$\therefore$ Component waves are,
and
$$ \begin{aligned} & y _1=2 \sin \left(\frac{\pi}{15} x-96 \pi t\right) \\ & y _2=2 \sin \left(\frac{\pi}{15} x+96 \pi t\right) \end{aligned} $$
BETA
