Wave Motion Ques 63
- A granite rod of $60 cm$ length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7 \times 10^{3} kg / m^{3}$ and its Young’s modulus is $9.27 \times 10^{10} Pa$. What will be the fundamental frequency of the longitudinal vibrations?
(2018 Main)
(a) $7.5 kHz$
(b) $5 kHz$
(c) $2.5 kHz$
(d) $10 kHz$
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Answer:
Correct Answer: 63.(b)
Solution:
Formula:
Vibrations of Strings (Standing Wave):
- Wave velocity $(v)=\sqrt{\frac{Y}{\rho}}=5.86 \times 10^{3} m / s$
For fundamental mode, $\lambda=2 l=1.2 m$
$\therefore \quad$ Fundamental frequency $=\frac{v}{\lambda}=4.88 kHz \approx 5 kHz$
BETA
