Stationary Waves & Vibrating Strings question from JEE Main 2002 (AIEEE)

Answer: (b)

1. A standing wave needs a wave of the same $a$, $\omega$, $k$ moving in the opposite direction: $y_2 = \pm a\sin(\omega t + kx)$.
2. Node at $x = 0$: $y_1 + y_2 = 0$ for all $t$, i.e. $a\sin\omega t + y_2(0, t) = 0$.
3. This needs $y_2 = -a\sin(\omega t + kx)$; then $y = -2a\cos\omega t\sin kx$, which is zero at $x = 0$.
4. So the answer is (b) $y = -a\sin(\omega t + kx)$.