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Question: A tuning fork of frequency \(220\,Hz\) produces sound waves of wavelength \(1.5\,m\) in air at S.T.P...

A tuning fork of frequency 220Hz220\,Hz produces sound waves of wavelength 1.5m1.5\,m in air at S.T.P. Calculate the increase in the wavelength, when the temperature of air is 27C27^\circ C.

Explanation

Solution

in order to solve the question, we will first find the velocity at temperature zero degree using the relation of frequency and wavelength then we will use the relation of velocity and temperature to find the velocity at twenty-seven degrees after that using the relation of velocity and wavelength we will find the wavelength at twenty-seven degrees then we will find the difference in both the wavelength

Formula used:
v=λfv = \lambda f
Here, vv is the velocity of wave, ff is the frequency of wave and λ\lambda is the wavelength.
vTv \propto \sqrt T
Here, vv is the velocity and TT is the temperature.

Complete step by step answer:
In the question we are given that a tuning fork produces sound in air at S.T.P and we have to find the increase in the wavelength, when the temperature of air is 27C27^\circ C.
Frequency of tuning fork = 220 Hz and Wavelength of sound waves = 1.5m.Now we will find the velocity with the help of the relation of frequency and wavelength at T=0C=273KT = 0^\circ C = 273K.
v1=λ1f\Rightarrow {v_1} = {\lambda _1}f
f=220Hz\Rightarrow f= 220\,Hz
λ\Rightarrow \lambda = 1.5m
Substituting the values in the formula
v=1.5×220v = 1.5 \times 220
v1=330 Hz\Rightarrow {v_1} = 330{\text{ Hz}}
Now we will use the relation of temperature to find the velocity at 27C27^\circ C
vTv \propto \sqrt T which means vT=constant\dfrac{v}{{\sqrt T }} = {\text{constant}}

Let the velocity at 27C27^\circ Cbe v2{v_2}.Therefore,
v1T1=v2T2\dfrac{{{v_1}}}{{\sqrt {{T_1}} }} = \dfrac{{{v_2}}}{{\sqrt {{T_2}} }}
v2=v1T2T1\Rightarrow {v_2} = {v_1}\sqrt {\dfrac{{{T_2}}}{{{T_1}}}}
Substituting the values
v2=330300273{v_2} = 330\sqrt {\dfrac{{300}}{{273}}}
v2=345.9 ms1\Rightarrow {v_2} = 345.9{\text{ }}m{s^{ - 1}}
Now we will find the velocity with the help of relation of frequency and wavelength T=27C=300KT = 27^\circ C = 300K.
v2=λ2f{v_2} = {\lambda _2}f
Substituting the value
λ2=345.9220{\lambda _2} = \dfrac{{345.9}}{{220}}
λ2=1.57m\Rightarrow {\lambda _2} = 1.57m
Change in wavelength
Δλ=λ2λ1\Delta \lambda = {\lambda _2} - {\lambda _1}
Δλ=1.571.5\Rightarrow \Delta \lambda = 1.57 - 1.5
Δλ=0.07m\therefore \Delta \lambda = 0.07\,m

Hence, the answer is Δλ=0.07m\Delta \lambda = 0.07\,m.

Note: Many students will make the mistake by not using the relation of the velocity instead of that using the whole formula which consist of mass, gas constant, temperature etc. but instead using we will use the given variable in the question and will be assuming all the other component constant which will give us the desired relation