Question
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?
✓
Answer
Wire of sonometer is twice the length which it vibrates in its second harmonic. Thus, if the tuning fork resonates at L, it will resonate at 2L. This can be explained as below: The frequency of sonometer is given by$\text{f}=\frac{\text{n}}{2\text{L}}\sqrt{\frac{\text{T}}{\mu}}=\frac{\text{nv}}{2\text{L}}$ (n = number of loops)
For a given sonometer velocity of wave will be constant. if after chaning the leggth of wire the tuing fork still be in resonance witrh the wire. then, $\frac{\text{n}}{\text{L}}=\text{constant}\Rightarrow\frac{\text{n}^2}{\text{L}^2}$$\frac{\text{n}^1}{\text{L}^1}=\frac{\text{n}^2}{2\text{L}^2}\Rightarrow\text{n}_2=2\text{n}_1$
Hence, when the wire is doubled the number of loops also get doubled to produce the resonance. That is it resonates in second harmonic.
For a given sonometer velocity of wave will be constant. if after chaning the leggth of wire the tuing fork still be in resonance witrh the wire. then, $\frac{\text{n}}{\text{L}}=\text{constant}\Rightarrow\frac{\text{n}^2}{\text{L}^2}$$\frac{\text{n}^1}{\text{L}^1}=\frac{\text{n}^2}{2\text{L}^2}\Rightarrow\text{n}_2=2\text{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.
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