The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is $360 \mathrm{~ms}^{-1}$ and their frequencies are 256 Hz .
(a) Calculate the time at which the second curve is plotted.
(b) Mark nodes and antinodes on the curve.
(c) Calculate the distance between $A^{\prime}$ and $C^{\prime}$.
Given, frequency of the wave $v=256 \mathrm{~Hz}$
Time period $$T=\frac{1}{v}=\frac{1}{256} \mathrm{~s}=3.9 \times 10^{-3} \mathrm{~s}$$
$$\begin{aligned} &\text { (a) Time taken to pass through mean position is }\\ &t=\frac{T}{4}=\frac{1}{40}=\frac{3.9 \times 10^{-3}}{4} \mathrm{~s}=9.8 \times 10^{-4} \mathrm{~s} \end{aligned}$$
(b) Nodes are $A, B, C, D, E$ (i.e., zero displacement)
Antinodes are $A^{\prime}, C^{\prime}$ (i.e., maximum displacement)
(c) It is clear from the diagram $A^{\prime}$ and $C^{\prime}$ are consecutive antinodes, hence separation $=$ wavelength $(\lambda)$
$$=\frac{v}{v}=\frac{360}{256}=1.41 \mathrm{~m}\quad$$ $$[\therefore v=v \lambda]$$
A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water (figure). The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, maximum intensity of sound is heard. If the room temperature is $20^{\circ} \mathrm{C}$, calculate
(a) speed of sound in air at room temperature.
(b) speed of sound in air at $0^{\circ} \mathrm{C}$.
(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?
Consider the diagram frequency of tuning fork $$\nu=512$$ Hz.
For observation of first maxima of intensity
(a) $L=\frac{\lambda}{4} \Rightarrow \lambda=4 L$ [for closed pipe]
$$\begin{aligned} v & =v \lambda=512 \times 4 \times 17 \times 10^{-2} \\ & =348.16 \mathrm{~m} / \mathrm{s} \end{aligned}$$
(b) We know that $v \propto \sqrt{T}$ where temperature $(T)$ is in kelvin.
$$\begin{aligned} & \frac{v_{20}}{v_0}=\sqrt{\frac{273+20}{273+0}}=\sqrt{\frac{293}{273}} \\ & \frac{v_{20}}{v_0}=\sqrt{1.073}=1.03 \\ & v_0=\frac{v_{20}}{1.03}=\frac{348.16}{1.03}=338 \mathrm{~m} / \mathrm{s} \end{aligned}$$
(c) Resonance will be observed at 17 cm length of air column, only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface because mercury is more denser than water.
Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio $1: 2: 3: 4$.
Let, there are n number of loops in the string.
Length corresponding each loop is $\frac{\lambda}{2}$.
Now, we can write
$$\begin{aligned} &L=\frac{n \lambda}{2} \Rightarrow \lambda=\frac{2 L}{n}\quad\text { [for } n \text { loops] } \end{aligned}$$
$$\begin{array}{ll} \Rightarrow & \frac{v}{v}=\frac{2 L}{n} \Rightarrow[\because v=v \lambda] \\ \Rightarrow & v=\frac{n}{2 L} v=\frac{n}{2 L} \sqrt{\frac{T}{\mu}} \quad [\because \text { velocity of transverse waves }=\sqrt{T / \mu}] \end{array}$$
$$\begin{aligned} &\Rightarrow v \propto n\quad [\because \text { length and speed are constants] } \end{aligned}$$
$$\begin{aligned} &\text { So, }\\ &\begin{aligned} v_1: v_2: v_3: v_4 & =n_1: n_2: n_3: n_4 \\ & =1: 2: 3: 4 \end{aligned} \end{aligned}$$
The earth has a radius of 6400 km . The inner core of 1000 km radius is solid. Outside it, there is a region from 1000 km to a radius of 3500 km which is in molten state. Then again from 3500 km to 6400 km the earth is solid. Only longitudinal $(P)$ waves can travel inside a liquid. Assume that the $P$ wave has a speed of $8 \mathrm{~km} \mathrm{~s}^{-1}$ in solid parts and of $5 \mathrm{~km} \mathrm{~s}^{-1}$ in liquid parts of the earth. An earthquake occurs at some place close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth, if wave travels along diameter?
Speed of wave in solid = 8 km/s
$$\begin{aligned} &\text { Speed of wave in liquid }=5 \mathrm{~km} / \mathrm{s}\\ &\text { Required time }=\left[\frac{1000-0}{8}+\frac{3500-1000}{5}+\frac{6400-3500}{8}\right] \times 2 \quad[\because \text { diameter }=\text { radius } \times 2] \end{aligned}$$
$$=\left[\frac{1000}{8}+\frac{2500}{5}+\frac{2900}{8}\right] \times 2 \quad\left[\text { time }=\frac{\text { distance }}{\text { speed }}\right]$$
$$=[125+500+362.5] \times 2=1975$$
As we are considering at diametrically opposite point, hence there is a multiplication of 2.
If $c$ is rms speed of molecules in a gas and $v$ is the speed of sound waves in the gas, show that $c / v$ is constant and independent of temperature for all diatomic gases.
We know that rms speed of molecules of a gas
$$c=\sqrt{\frac{3 p}{\rho}}=\sqrt{\frac{3 R T}{M}}\quad \text{.... (i)}$$
where $M=$ molar mass of the gas.
Speed of sound wave in gas $v=\sqrt{\frac{\gamma p}{\rho}}=\sqrt{\frac{\gamma R T}{M}}\quad \text{... (ii)}$
On dividing Eq. (i) by Eq. (ii), we get
$$\frac{c}{v}=\sqrt{\frac{3 R T}{M} \times \frac{M}{\gamma R T}} \Rightarrow \frac{c}{v}=\sqrt{\frac{3}{\gamma}}$$
where $\gamma=$ adiabatic constant for diatomic gas
$$\gamma=\frac{7}{5}\quad \left[\text { since } \gamma=\frac{C_p}{C_V}\right]$$
Hence, $$\frac{c}{v}=\text { constant }$$