JEE Advanced 2014 · Paper 1 · Physics Q3

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PhysicsMultiple correct+3Question 3 of 60

One end of a taut string of length 3 m along the $x$ axis is fixed at $x=0$ . The speed of the waves in the string is $100\ \mathrm{m\,s^{-1}}$ . The other end of the string is vibrating in the $y$ direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is(are)

y(t)=A\sin\dfrac{\pi x}{6}\cos\dfrac{50\pi t}{3}

y(t)=A\sin\dfrac{\pi x}{3}\cos\dfrac{100\pi t}{3}

y(t)=A\sin\dfrac{5\pi x}{6}\cos\dfrac{250\pi t}{3}

y(t)=A\sin\dfrac{5\pi x}{2}\cos250\pi t

Answer and worked solution

Official answer: A, C, D

Boundary conditions. $x=0$ is fixed (a node), and the driven end $x=3$ m behaves as an antinode. So
$L=(2n+1)\frac\lambda4\implies k=\frac{2\pi}\lambda=\frac{(2n+1)\pi}{2L}=\frac{(2n+1)\pi}6,\qquad \omega=vk=100k.$

(A) $k=\frac\pi6$ ( $2n+1=1$ ), $\omega=\frac{100\pi}6=\frac{50\pi}3$ . Allowed.
(B) $k=\frac\pi3=\frac{2\pi}6$ : an even multiple, which puts a node at $x=3$ . Not allowed.
(C) $k=\frac{5\pi}6$ , $\omega=\frac{500\pi}6=\frac{250\pi}3$ . Allowed.
(D) $k=\frac{5\pi}2=\frac{15\pi}6$ , $\omega=250\pi$ . Allowed.

(A), (C), (D)

Topics:Standing waves in strings

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JEE Advanced 2014 · Paper 1, Physics question 3