JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Waves & Sound

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A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is

320 ms^{- 1}

, the mass of the string is

(A)5 grams
(B)10 grams✓ Correct
(C)20 grams
(D)40 grams

Answer: (B)

Solution

For a pipe closed at one end, the fundamental frequency is:

f_{p} = \frac{v _{s}}{4 L _{p}} = \frac{320}{4 \times 0.8} = 100 Hz

For a string vibrating in its second harmonic:

f_{s} = \frac{2}{2 ℓ _{s}} \frac{T}{μ} = \frac{1}{ℓ _{s}} \frac{T}{μ}

Since the frequencies are equal:

\frac{v _{s}}{4 L _{p}} = \frac{1}{ℓ _{s}} \frac{T}{μ}

Solving for

μ ℓ_{s}

(which is the mass of the string):

μ ℓ_{s} = \frac{T \times 4 L _{p}^{2}}{v _{s}^{2} \times ℓ _{s}} = \frac{50 \times 4 \times ( 0.8 ) ^{2}}{( 320 ) ^{2} \times 0.5} = 0.01 kg = 10 grams

The answer is (B) 10 grams.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Waves & Sound

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