JEE Advanced · 2006 · 09 Apr 2006 · Waves & Sound

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A massless rod is suspended by two identical strings AB and CD of equal length. A block of mass m is suspended from point O such that BO is equal to 'x'. Further, it is observed that the frequency of

1^{st}

harmonic (fundamental frequency) in AB is equal to

2^{nd}

harmonic frequency in CD. Then, length of BO is

(A)L/5✓ Correct
(B)4L/5
(C)3L/4
(D)L/4

Answer: (A)

Solution

For a string of length

ℓ

under tension T with linear mass density

μ

, the fundamental frequency is

f_{1} = \frac{1}{2 ℓ} \frac{T}{μ}

and the second harmonic is

f_{2} = \frac{1}{ℓ} \frac{T}{μ}

.

Given that the fundamental frequency in AB equals the second harmonic in CD:

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

This gives

T_{1} = 4 T_{2}

.

For rotational equilibrium about the suspension point, taking moments:

T_{1} \cdot x = T_{2} \cdot (L - x)

4 T_{2} \cdot x = T_{2} \cdot (L - x)

4 x = L - x

5 x = L

x = \frac{L}{5}

Therefore, the answer is (A).

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option

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JEE Advanced · 2006 · 09 Apr 2006 · Waves & Sound

Solution

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