JEE MAIN - Physics (2018 (Offline) - No. 1)
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 $$\times$$ 103 kg/m3 and its Young’s modulus is 9.27 $$\times$$ 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?
7.5 kHz
5 kHz
2.5 kHz
10 kHz
Explanation
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As rod length = 60 cm
$$\therefore\,\,\,$$ $${\lambda \over 2}$$ = 60
$$ \Rightarrow $$$$\,\,\,$$ $$\lambda $$ = 120 cm = 1.2 m
In solid, velocity of wave,
V = $$\sqrt {{Y \over \rho }} $$
= $$\sqrt {{{9.27 \times {{10}^{10}}} \over {2.7 \times {{10}^3}}}} $$
= 5.85 $$ \times $$ 103 m/sec.
As we know,
v = f $$\lambda $$
$$\therefore\,\,\,$$ f = $${v \over \lambda }$$
= $${{5.85 \times {{10}^3}} \over {1.2}}$$
= 4.88 $$ \times $$ 103 Hz
$$ \simeq $$ 5 kHz
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