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

In case of a string fixed at two ends, when the string vibrates in n loops
\({ v }_{ n }=\frac { n }{ 2l } \sqrt { \frac { T }{ \mu } } \quad \Rightarrow { v }_{ n }\propto n\)
Hence, when the string vibrates in 1 loop, 2 loops, 3 loops, 4 loops, the frequencies are in the ratio 1:2:3:4.

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
\(=\frac { v\times V }{ v-{ v }_{ s } } =\frac { 330\times 400 }{ 330-10 } \)
\(=412.5\ Hz\)

Number of beats/s = (n₁ - n₂)
Hence, time interval between two successive beats = time interval between two successive maxima = \(\frac { 1 }{ { n }_{ 1 }-{ n }_{ 2 } } \) 

The sonometer frequency is given by
\(v=\frac { n }{ 2L } \sqrt { \frac { T }{ \mu } } \)
Now, as it vibrates with length L, we assume v=v₁
n = n₁
\(\therefore { v }_{ 1 }=\frac { { n }_{ 1 } }{ 2L } \sqrt { \frac { T }{ \mu } } \)
When length is doubled, then
\({ v }_{ 2 }=\frac { { n }_{ 2 } }{ 2\times 2L } \sqrt { \frac { T }{ \mu } } \)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac { { v }_{ 1 } }{ { v }_{ 2 } } =\frac { { n }_{ 1 } }{ { n }_{ 2 } } \times 2\)
To keep the resonance,
\( \frac { { v }_{ 1 } }{ { v }_{ 2 } } =1=\frac { { n }_{ 1 } }{ { n }_{ 2 } } \times 2\)
\(\\ \Rightarrow { n }_{ 2 }=2{ n }_{ 1 }\)
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.