The sonometer frequency is given by
v=\(\frac{n}{2L}\sqrt{\frac{T}{μ}}\)
Now, as it vibrates with length L, we assume v=v₁
n = n₁
∴v₁=\(\frac{n_1}{2L}\sqrt{\frac{T}{μ}}\)
When length is doubled, then
v₂=\(\frac{n_2}{2 \times2L}\sqrt{\frac{T}{μ}}\)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac{v_1}{v_2}=\frac{n_1}{n_2}×2\)
To keep the resonance,
\(\frac{v_1}{v_2}=1=\frac{n_1}{n_2}×2\)
⇒n₂=2n₁
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.

Number of beats/s = (n₁ - n₂)
Hence, time interval between two successive beats = time interval between two successive maxima = \(\frac{1}{n_1−n_2}\)

Given, v=400 Hz, V_s = 10m/s, v = 330m/s
As the source is moving towards stationary observer, apparent frequency will be more than the original.
V^'=apparent frequency
\(=\frac{v×V}{v−v_s}=\frac{330×400}{330−10}\)
=412.5 Hz

In case of a string fixed at two ends, when the string vibrates in n loops
v_n= \(\frac{n}{2l}\sqrt{\frac{T}{μ}}\) ⇒v_n ∝ n
Hence, when the string vibrates in 1 loop, 2 loops, 3 loops, 4 loops, the frequencies are in the ratio 1:2:3:4.

Speed of sound waves in air increases with increases with increase in humidity. This is because presence of moisture decreases the density of air.