A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of $10 \mathrm{~m} / \mathrm{s}$. Given that the speed of sound in still air is $340 \mathrm{~m} / \mathrm{s}$. Then
Which of the following statement are true for a stationary waves?
A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire?
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.
An organ pipe of length $L$ open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of 480 Hz . What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?
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
$$\begin{array}{ll} \text { Hence, } & v=v^{\prime} \quad \text{[for resonance with same tuning fork]}\\ \Rightarrow & \frac{c}{2 L}=\frac{c}{4 L^{\prime}} \quad [\because \text{speed remains constant]}\\ \Rightarrow & \frac{L^{\prime}}{L}=\frac{2}{4}=\frac{1}{2} \Rightarrow L^{\prime}=\frac{L}{2} \end{array}$$
A tuning fork $A$, marked 512 Hz , produces 5 beats per second, where sounded with another unmarked tuning fork $B$. If $B$ is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork $B$ when not loaded?
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 $v_B=517 \mathrm{~Hz}$