Question

The ratio of displacement amplitude of wave at observation point to wavelength of sound waves is

• A. $3.22\times {10}^{-6}$
• B. $3.22\times {10}^{-12}$
• C. $3.22\times {10}^{-9}$
• D. $1.07\times {10}^{-10}$

Answer 1

Answer: (C)

$3.22\times {10}^{-9}$

As pressure oscillation amplitude $P_o=3\times {10}^{-3}N/m^2$ (solved earlier)

Let $s_o$ and $\lambda$ be the displacement amplitude and

wavelength of the wave respectively.

Now $P_o=Bk{s}_o$ where $B=p{v}^2$ and $k=\frac{2\pi }{\lambda }$

$⟹\frac{s_o}{\lambda }=\frac{P_o}{2\pi p{v}^2}=\frac{3\times {10}^{-3}}{2\times 3.14\times (330{)}^2\times \frac{15}{11}}$

$⟹\frac{s_o}{\lambda }=3.22\times {10}^{-9}$