Wave Motion 2 Question 9

9. 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) $10 kHz$

Show Answer

Answer:

Correct Answer: 9. (b)

Solution:

  1. 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 $=\frac{v}{\lambda}=4.88 kHz \approx 5 kHz$



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक