JEE Advanced 2023 Paper 2 · Physics · Waves and Sound · easy

A string of length 1.20 m is fixed at one end and its other end is driven by a tuning fork of constant frequency (treat both ends as nodes). The…

Numerical answer

A string of length

1.20\ \mathrm{m}

is fixed at one end and its other end is driven by a tuning fork of constant frequency (treat both ends as nodes). The string is kept taut by passing it over a light frictionless pulley and attaching a block of density

1600\ \mathrm{kg\,m^{-3}}

to it. With the block hanging in air the string resonates in its

3^{\mathrm{rd}}

harmonic. When the block is completely immersed in a liquid of density

700\ \mathrm{kg\,m^{-3}}

(the string stays out of the liquid) it again resonates with the same fork. Find the order

n

of the harmonic in which it now vibrates.

Show answer and solution

Answer: 4

With the fork frequency

f

fixed,

f=\dfrac{n}{2L}\sqrt{T/\mu}

, so

n\sqrt{T}

is constant. In air the tension is the weight

mg

. In the liquid, buoyancy reduces it to

T'=mg\left(1-\dfrac{\rho_\ell}{\rho_b}\right)=mg\left(1-\dfrac{700}{1600}\right)=\dfrac{9}{16}\,mg

. Hence

\sqrt{T'/T}=\dfrac34

and

n'=n\sqrt{T/T'}=3\times\dfrac43=4

. The string therefore vibrates in the

4^{\mathrm{th}}

harmonic.

Report a mistake

← Previous Waves and Sound question Next Waves and Sound question →

Original question written in the JEE Advanced 2023 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake