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 $v=330 \mathrm{~m} / \mathrm{s}$
$$\text { Apparent frequency when source is moving } v_{\mathrm{app}}=\left(\frac{v}{v-v_t}\right) v$$
$$\begin{aligned} = & \left(\frac{330}{330-10}\right) 400 \\ \Rightarrow \quad v_{\text {app }} & =\frac{330}{320} \times 400=412.5 \mathrm{~Hz} \end{aligned}$$
The wave pattern on a stretched string is shown in figure. Interpret what kind of wave this is and find its wavelength.
We have to observe the displacement and position of different points, then accordingly nature of two wave is decided.
Points on positions $x=10,20,30,40$ never move, always at mean position with respect to time. These are forming nodes which characterise a stationary wave.
$$\begin{aligned} & \because \text { Distance between two successive nodes }=\frac{\lambda}{2} \\ & \begin{aligned} \Rightarrow \quad \lambda & =2 \times(\text { node to node distance }) \\ & =2 \times(20-10) \\ & =2 \times 10=20 \mathrm{~cm} \end{aligned} \end{aligned}$$
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
$$\begin{aligned} &L=\frac{n \lambda}{2} \Rightarrow \lambda=\frac{2 L}{n}\quad\text { [for } n \text { loops] } \end{aligned}$$
$$\begin{array}{ll} \Rightarrow & \frac{v}{v}=\frac{2 L}{n} \Rightarrow[\because v=v \lambda] \\ \Rightarrow & v=\frac{n}{2 L} v=\frac{n}{2 L} \sqrt{\frac{T}{\mu}} \quad [\because \text { velocity of transverse waves }=\sqrt{T / \mu}] \end{array}$$
$$\begin{aligned} &\Rightarrow v \propto n\quad [\because \text { length and speed are constants] } \end{aligned}$$
$$\begin{aligned} &\text { So, }\\ &\begin{aligned} v_1: v_2: v_3: v_4 & =n_1: n_2: n_3: n_4 \\ & =1: 2: 3: 4 \end{aligned} \end{aligned}$$