A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?
$$\begin{aligned} &\text { Frequency of vibrations produced by a stretched wire }\\ &\mathrm{v}=\frac{n}{2 l} \sqrt{\frac{T}{\mu}} \end{aligned}$$
$$\text { Mass per unit length } \mu=\frac{\text { Mass }}{\text { Length }}=\frac{\pi r^2 l \rho}{l}=\pi r^2 \rho \quad\left[\because M=v p=A l \rho=\pi r^2 l \rho\right]$$
$$\begin{aligned} &\begin{aligned} \therefore \quad v & =\frac{n}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}} \Rightarrow \mathrm{v} \propto \sqrt{\frac{1}{r^2}} \\ & \mathrm{v} \propto \frac{1}{r} \end{aligned}\\ &\text { Hence, when radius is tripled, } v \text { will be } \frac{1}{3} \text { rd of previous value. } \end{aligned}$$
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.