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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Payal Gupta | Contributor-Level 10

3 months ago

This is a short answer type question as classified in NCERT Exemplar

Wire of twice the length vibrates in its second harmonic . thus, if the tuning fork resonates at L, it will resonate at 2L

So the sonometer frequency is

$v = \frac{n}{2 l} \sqrt[]{\frac{T}{m}}$

Now if it vibrates with length L we assume $v = v_{1}$

n=n₁

$v = \frac{n}{2 L} \sqrt[]{\frac{T}{m}}$

When length is doubled then

$v_{2} = \frac{n_{1}}{2 \times 2 L} \sqrt[]{\frac{T}{m}}$

Dividing above equations

$\frac{v_{1}}{v_{2}} = \frac{n_{1}}{n_{2}} \times 2$

To Keep the resonance $\frac{v_{1}}{v_{2}} = 1 = \frac{n_{1}}{n_{2}} \times 2$

n₂=2n₁ so it resonates 2^nd harmonic.

This is a short answer type question as classified in NCERT ExemplarWire of twice the length vibrates in its second harmonic . thus, if the tuning fork resonates at L, it will resonate at 2LSo the sonometer frequency is <math> <mi>v</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>l</mi> </mrow> </mrow> </mfrac> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> Now if it vibrates with length L we assume <math> <mi>v</mi> <mo>=</mo> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> n=n1 <math> <mi>v</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>L</mi> </mrow> </mrow> </mfrac> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> When length is doubled then <math> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mo>×</mo> <mn>2</mn> <mi>L</mi> </mrow> </mrow> </mfrac> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> Dividing above equations <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>×</mo> <mn>2</mn> </math> To Keep the resonance <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>×</mo> <mn>2</mn> </math> n2=2n1 so it resonates 2nd harmonic.

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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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