JEE Advanced 2013 · Paper 1 · Physics Q11

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PhysicsMultiple correct+4 / -1Question 11 of 60

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y(x,t)=(0.01\ \mathrm m)\sin\left[\left(62.8\ \mathrm m^{-1}\right)x\right]\cos\left[\left(628\ \mathrm s^{-1}\right)t\right]$ . Assuming $\pi=3.14$ , the correct statement(s) is (are)

The number of nodes is 5.

The length of the string is 0.25 m.

The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.

The fundamental frequency is 100 Hz.

Answer and worked solution

Official answer: B, C

Wavelength. $k=62.8=20\pi$ m⁻¹, so $\lambda=0.1$ m.

(B) Fifth harmonic: $L=\frac{5\lambda}2=0.25$ m. ✓
(A) Five loops have 6 nodes (counting the ends). ✗
(C) At the midpoint $x=0.125$ m, $\sin(20\pi\times0.125)=\sin2.5\pi=1$ : an antinode, amplitude 0.01 m. ✓
(D) $\omega=628$ s⁻¹ gives 100 Hz, the fifth harmonic; the fundamental is 20 Hz. ✗

(B), (C)

Topics:Standing waves in strings

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JEE Advanced 2013 · Paper 1, Physics question 11