A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y(x, t)=(0.01 m) \sin \left[\left(62.8 m^{-1}\right) x\right] \cos \left[\left(628 s^{-1}\right) t\right]$. Assuming $\pi=3.14$, the correct statement(s) is (are) (A) The number of nodes is 5. (B) The length of the string is 0.25 m . (C) The maximum displacement of the midpoint of the string, from its equilibrium position is $0.01 m$. (D) The fundamental frequency is 100 Hz .

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Waves & Sound

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,

y (x, t) = (0.01 m) sin [(62.8 m^{- 1}) x] cos [(628 s^{- 1}) t]

. Assuming

π = 3.14

, the correct statement(s) is (are)
(A) The number of nodes is 5.
(B) The length of the string is 0.25 m .
(C) The maximum displacement of the midpoint of the string, from its equilibrium position is

0.01 m

.
(D) The fundamental frequency is 100 Hz .

(A)The number of nodes is 5.
(B)The length of the string is 0.25 m .✓ Correct
(C)The maximum displacement of the midpoint of the string, from its equilibrium position is $0.01 m$ .✓ Correct
(D)The fundamental frequency is 100 Hz .

Answer: (B), (C)

Solution

From the wave equation

y (x, t) = 0.01 sin (62.8 x) cos (628 t)

:

Wave number

k = 62.8 = \frac{2 π}{λ}

λ = \frac{2 \times 3.14}{62.8} = \frac{1}{10} m

For the fifth harmonic:

L = \frac{5 λ}{2} = \frac{5}{2} \times \frac{1}{10} = \frac{1}{4} = 0.25 m

✓ (B is correct)

The number of nodes in the fifth harmonic is 6 (not 5), so (A) is incorrect.

At the midpoint (antinode), the maximum displacement is the amplitude = 0.01 m ✓ (C is correct)

Wave velocity

v = \frac{ω}{k} = \frac{628}{62.8} = 10 m / s

Fundamental frequency

f_{1} = \frac{v}{2 L} = \frac{10}{2 \times 0.25} = 20 H z

(not 100 Hz), so (D) is incorrect.

Hence the correct answers are (B) and (C).

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Waves & Sound

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