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NCERT Exemplar · Q16

Q.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?

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If the wire originally vibrates in its nn-th harmonic to match the tuning fork, doubling the wire's length (tension and linear density unchanged) means the fork will resonate again only if the wire now vibrates in exactly twice its original harmonic (loop) number, i.e. m=2nm=2n.

Setting up the resonance condition

For a stretched wire fixed at both ends, the kk-th harmonic frequency is:

fk=k2LTμf_k = \frac{k}{2L}\sqrt{\frac T\mu}

Initial resonance. Suppose the wire, at its original length LL, resonates with the tuning fork while vibrating in its nn-th harmonic:

ffork=fn=n2LTμf_{\text{fork}} = f_n = \frac{n}{2L}\sqrt{\frac T\mu}

After doubling the length. With the new length L′=2LL'=2L (tension TT and μ\mu unchanged), the wire's mm-th harmonic frequency becomes:

fm′=m2L′Tμ=m4LTμf_m' = \frac{m}{2L'}\sqrt{\frac T\mu} = \frac{m}{4L}\sqrt{\frac T\mu}

Finding the condition for continued resonance

For the tuning fork to still be in resonance, we need fm′=fnf_m' = f_n:

m4LTμ=n2LTμ\frac{m}{4L}\sqrt{\frac T\mu} = \frac{n}{2L}\sqrt{\frac T\mu}

The common factor cancels:

m4=n2⇒m=2n\frac m4 = \frac n2 \quad\Rightarrow\quad m = 2n

Interpretation

Whatever harmonic the wire originally vibrated in (nn), it must vibrate in exactly double that harmonic number after the length is doubled, to reproduce the original frequency. This makes sense: doubling the length halves the fundamental frequency, so to recover the original frequency you need a harmonic number twice as large. …

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