Wave Motion Ques 65
- A steel wire of length $1 m$, mass $0.1 kg$ and uniform cross-sectional area $10^{-6} m^{2}$ is rigidly fixed at both ends. The temperature of the wire is lowered by $20^{\circ} C$. If transverse waves are set-up by plucking the string in the middle, calculate the frequency of the fundamental mode of vibration.
Given :
$Y _{\text {steel }}=2 \times 10^{11} N / m^{2}$,
$ \alpha _{\text {steel }}=1.21 \times 10^{-5} /{ }^{\circ} C \text {. } $
$(1984,6 M)$
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Answer:
Correct Answer: 65.$11 Hz$
Solution:
Formula:
Vibrations of Strings (Standing Wave):
- The temperature stress is $\sigma=Y \alpha \Delta \theta$ or tension in the steel wire $T=\sigma A=Y A \alpha \Delta \theta$ Substituting the values, we have
$$ T=\left(2 \times 10^{11}\right)\left(10^{-6}\right)\left(1.21 \times 10^{-5}\right)(20)=48.4 N $$
Speed of transverse wave on the wire, $v=\sqrt{\frac{T}{\mu}}$
Here, $\mu=$ mass per unit length of wire $=0.1 kg / m$
$$ \therefore \quad v=\sqrt{\frac{48.4}{0.1}}=22 m / s $$
Fundamental frequency
$$ f _o=\frac{v}{2 l}=\frac{22}{2 \times 1}=11 Hz $$
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