At what temperatures (in ${ }^{\circ} \mathrm{C}$ ) will the speed of sound in air be 3 times its value at $0^{\circ} \mathrm{C}$ ?
We know that speed of sound in air $v \propto \sqrt{T}$
$\therefore \frac{v_T}{v_0}=\sqrt{\frac{T_T}{T_0}}=\sqrt{\frac{T_T}{273}}\quad$ [where $T$ is in kelvin]
$$\begin{aligned} \text{But}\quad & \frac{v_T}{v_0}=\frac{3}{1} \quad [\because \text{ speed becomes three times]}\\ \therefore\quad & \frac{3}{1}=\sqrt{\frac{T_T}{T_0}} \Rightarrow \frac{T_T}{273}=9 \\ \therefore \quad &T_T=273 \times 9=2457 \mathrm{~K} \\ & =2457-273=2184^{\circ} \mathrm{C} \end{aligned}$$
When two waves of almost equal frequencies $n_1$ and $n_2$ reach at a point simultaneously, what is the time interval between successive maxima?
Let, $$n_1>n_2$$
Beat frequency
$$\begin{aligned} & \quad v_b=n_1-n_2 \\ & \therefore \quad \text { Time period of beats }=T_b=\frac{1}{v_b}=\frac{1}{n_1-n_2} \end{aligned}$$
A steel wire has a length of 12 m and a mass of 2.10 kg . What will be the speed of a transverse wave on this wire when a tension of $2.06 \times 10^4 \mathrm{~N}$ is applied?
$$\begin{aligned} &\text { Given, length of the wire }\\ &l=12 \mathrm{~m} \end{aligned}$$
Mass of wire $$m=2.10 \mathrm{~kg}$$
Tension $T=2.06 \times 10^4 \mathrm{~N}$
Speed of transverse wave $\quad v=\sqrt{\frac{T}{\mu}}$ [where $\mu=$ mass per unit length]
$$=\sqrt{\frac{2.06 \times 10^4}{\left(\frac{2.10}{12}\right)}}=\sqrt{\frac{2.06 \times 12 \times 10^4}{2.10}}=343 \mathrm{~m} / \mathrm{s}$$
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz ? (sound velocity in air $=330 \mathrm{~ms}^{-1})$
Length of pipe
$$\begin{aligned} & v_{\text {funda }}=\frac{V}{4 L}=\frac{330}{4 \times 20 \times 10^{-2}} \quad \text{(for closed pipe)}\\ & v_{\text {funda }}=\frac{330 \times 100}{80}=412.5 \mathrm{~Hz} \\ & \frac{v_{\text {given }}}{v_{\text {funda }}}=\frac{1237.5}{412.5}=3 \end{aligned}$$
Hence, 3rd harmonic node of the pipe is resonantly excited by the source of given frequency.
A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air. The train begins to move with a speed of $10 \mathrm{~ms}^{-1}$ towards the platform. What is the frequency of the sound for an observer standing on the platform? (sound velocity in air $=330 \mathrm{~ms}^{-1}$ )
As the source (train) is moving towards the observer (platform) hence apparent frequency observed is more than the natural frequency.
Frequency of whistle $\nu=400 \mathrm{~Hz}$
Speed of train $v_t=10 \mathrm{~m} / \mathrm{s}$
Velocity of sound in air $v=330 \mathrm{~m} / \mathrm{s}$
$$\text { Apparent frequency when source is moving } v_{\mathrm{app}}=\left(\frac{v}{v-v_t}\right) v$$
$$\begin{aligned} = & \left(\frac{330}{330-10}\right) 400 \\ \Rightarrow \quad v_{\text {app }} & =\frac{330}{320} \times 400=412.5 \mathrm{~Hz} \end{aligned}$$