QJEEM18Phddea: JEE Main 2018 Physics - Class 11 JEEM18Phddea Solution: A granite rod of 60 cm length is clamped at its middle point…

Question

Verified Solution • 2025–26 Session moderate

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 ?

Accepted Answer

As rod length = 60 cm

∴

\frac{λ}{2}

= 60

\Rightarrow

λ

= 120 cm = 1.2 m

In solid, velocity of wave,

V =

\sqrt{\frac{Y}{ρ}}

=

\sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^{3}}}

= 5.85

\times

10³ m/sec.

As we know,

v = f

λ

∴

f =

\frac{v}{λ}

=

\frac{5.85 \times 10^{3}}{1.2}

= 4.88

\times

10³ Hz

≃

5 kHz

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