A tuning fork makes $400$ vibrations in two seconds in air. If velocity of sound at $0 K$ is $320 m s^{- 1}$ then find the wavelength of the wave emitted at $27 K$ .

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Solution

Given , $T_{1} = 0 K = 0 + 273 = 273 K$ ,
$T_{2} = 27 K = 27 + 273 = 300 K$ ,
$v_{1} = 320 m / s$
We have , $v \propto \sqrt{T}$
Hence , $\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$
or $\frac{v_{2}}{320} = \sqrt{\frac{300}{273}}$
or $v_{2} = 320 \times 1.04 = 332.8 m / s$
Now , given , frequency of sound
$n = 400 / 2 = 200 H z$
Hence , wavelength of wave at 27K
$λ = v_{2} / n = 332.8 / 200 = 1.66 m$

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A tuning fork makes 400 vibrations in two seconds in air. If velocity of sound at 0K is 320ms−1 then find the wavelength of the wave emitted at 27K.

Given , T1=0K=0+273=273K , T2=27K=27+273=300K , v1=320m/s We have , v∝√THence , v2v1=√T2T1 or v2320=√300273or v2=320×1.04=332.8m/sNow , given , frequency of sound n=400/2=200HzHence , wavelength of wave at 27K λ=v2/n=332.8/200=1.66m

A tuning fork makes $400$ vibrations in two seconds in air. If velocity of sound at $0 K$ is $320 m s^{- 1}$ then find the wavelength of the wave emitted at $27 K$ .