Wire of twice the length vibrates in its second harmonic. Thus, if the tuning fork resonates at $L$, it will resonate at $2 L$. This can be explained as below The sonometer frequency is given by

$$\mathrm{v}=\frac{n}{2 L} \sqrt{\frac{T}{m}} \quad(n=\text { number of loops })$$

$$\begin{aligned} &\text { Now, as it vibrates with length } L \text {, we assume } v=v_1\\ &\begin{aligned} n & =n_1 \\ \therefore \quad v_1 & =\frac{n_1}{2 L} \sqrt{\frac{T}{m}} \quad \text{... (i)} \end{aligned}\end{aligned}$$

$$\begin{aligned} &\text { When length is doubled, then }\\ &v_2=\frac{n_2}{2 \times 2 L} \sqrt{\frac{T}{m}}\quad \text{.... (ii)} \end{aligned}$$

Dividing Eq. (i) by Eq. (ii), we get

$$\frac{v_1}{v_2}=\frac{n_1}{n_2} \times 2$$

To keep the resonance

$$\begin{aligned} & \frac{v_1}{v_2}=1=\frac{n_1}{n_2} \times 2 \\ & n_2=2 n_1 \end{aligned}$$

Hence, when the wire is doubled the number of loops also get doubled to produce the resonance. That is it resonates in second harmonic.

Consider the situation shown in the diagram

As the organ pipe is open at both ends, hence for first harmonic

$$l=\frac{\lambda}{2}$$

$$\Rightarrow \quad \lambda=2 l \Rightarrow \frac{c}{v}=2 l \Rightarrow v=\frac{c}{2 l}$$

where $c$ is speed of the sound wave in air.

For pipe closed at one end

$$v^{\prime}=\frac{C}{4 L^{\prime}}$$

c for first harmonic

$$\begin{array}{ll} \text { Hence, } & v=v^{\prime} \quad \text{[for resonance with same tuning fork]}\\ \Rightarrow & \frac{c}{2 L}=\frac{c}{4 L^{\prime}} \quad [\because \text{speed remains constant]}\\ \Rightarrow & \frac{L^{\prime}}{L}=\frac{2}{4}=\frac{1}{2} \Rightarrow L^{\prime}=\frac{L}{2} \end{array}$$

Frequency of tuning fork $A$,

$$v_A=512 \mathrm{~Hz}$$

Probable frequency of tuning fork $B$,

$$v_B=v_A \pm 5=512 \pm 5=517 \text { or } 507 \mathrm{~Hz}$$

when $B$ is loaded, its frequency reduces.

If it is 517 Hz , it might reduced to 507 Hz given again a beat of 5 Hz .

If it is 507 Hz , reduction will always increase the beat frequency, hence $v_B=517 \mathrm{~Hz}$

Given, displacement of an elastic wave $y=3 \sin \omega t+4 \cos \omega t$

Assume,

$$\begin{aligned} & 3=a \cos \phi \quad \text{... (i)}\\ & 4=a \sin \phi \quad \text{... (ii)} \end{aligned}$$

$$\begin{aligned} &\text { On dividing Eq. (ii) by Eq. (i) }\\ &\tan \phi=\frac{4}{3} \Rightarrow \phi=\tan ^{-1}(4 / 3) \end{aligned}$$

$$\begin{array}{ll} \text { Also, } & a^2 \cos ^2 \phi+a^2 \sin ^2 \phi=3^2+4^2 \\ \Rightarrow & a^2\left(\cos ^2 \phi+\sin ^2 \phi\right)=25 \end{array}$$

$$a^2 \cdot 1=25 \Rightarrow a=5$$

$$\begin{aligned} \text{Hence,}\quad Y & =5 \cos \phi \sin \omega t+5 \sin \phi \cos \omega t \\ & =5[\cos \phi \sin \omega t+\sin \phi \cos \omega t]=5 \sin (\omega t+\phi) \end{aligned}$$

where $\quad \phi=\tan ^{-1}(4 / 3)$

Hence, amplitude $=5 \mathrm{~cm}$

$$\begin{aligned} &\text { Frequency of vibrations produced by a stretched wire }\\ &\mathrm{v}=\frac{n}{2 l} \sqrt{\frac{T}{\mu}} \end{aligned}$$

$$\text { Mass per unit length } \mu=\frac{\text { Mass }}{\text { Length }}=\frac{\pi r^2 l \rho}{l}=\pi r^2 \rho \quad\left[\because M=v p=A l \rho=\pi r^2 l \rho\right]$$

$$\begin{aligned} &\begin{aligned} \therefore \quad v & =\frac{n}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}} \Rightarrow \mathrm{v} \propto \sqrt{\frac{1}{r^2}} \\ & \mathrm{v} \propto \frac{1}{r} \end{aligned}\\ &\text { Hence, when radius is tripled, } v \text { will be } \frac{1}{3} \text { rd of previous value. } \end{aligned}$$