Question

$40$ waves pass through a point in $0.5s$ at $0^{o}C$. The distance between one compression and adjacent rarefaction is $8m$. Find the velocity of the sound wave at $2457K$.

Answer 1

If $40$ waves pass through a point in $0.5s$ then it means one wave takes $\frac{0.5}{40}sec$

This means that the time period 'T' is $\frac{0.5}{40}sec$

The distance between one compression and adjacent rarefaction is half of wavelength. So wavelength $\lambda =16m$

We know that $speed=\frac{\lambda }{T}$

Thus the speed at $0^{0}Celcius=v_1=16÷\frac{0.5}{40}=\frac{16\times 40}{0.5}=1280$ m/s

From the given conditions we have $T_1=0^{0}Celcius={273}^{0}Kelvin$ and $T_2=2457K$

We know that $\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{2457}{273}}=3$

Hence $v_2=v_1\times 3=1280\times 3=3840$ m/s

The velocity of the sound wave at $2457K$ is $3840m/s$