Wave Motion Ques 48
- In a sonometer wire, the tension is maintained by suspending a $50.7 kg$ mass from the free end of the wire. The suspended mass has a volume of $0.0075 m^{3}$. The fundamental frequency of vibration of the wire is $260 Hz$. If the suspended mass is completely submerged in water, the fundamental frequency will become ….. Hz.
(1987, 2M)
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Answer:
Correct Answer: 48.240
Solution:
Formula:
Vibrations of Strings (Standing Wave):
- Fundamental frequency $f=\frac{v}{2 l}=\frac{\sqrt{T / \mu}}{2 l}$ or $f \propto \sqrt{T}$
$$ \therefore \quad \frac{f^{\prime}}{f}=\sqrt{\frac{w-F}{w}} $$
Here, $w=$ weight of mass and $F=$ upthrust
$$ \therefore \quad f^{\prime}=f \sqrt{\frac{w-F}{w}} $$
Substituting the values, we have
$$ f^{\prime}=260 \sqrt{\frac{(50.7) g-(0.0075)\left(10^{3}\right) g}{(50.7) g}}=240 Hz $$
BETA
