Q 11-14-119JEE MainJEE Main 2019 (9 Apr, Shift 1)Easy
A string is clamped at both the ends and it is vibrating in its $4^{\text{th}}$ harmonic. The equation of the stationary wave is $y = 0.3\sin(0.157x)\cos(200\pi t)$. The length of the string is (All quantities are in SI units.)
Answer: (D) $80\ \text{m}$
$k = 0.157 = \dfrac{\pi}{20}\ \text{m}^{-1}$, so $\lambda = \dfrac{2\pi}{k} = 40\ \text{m}$.
In the $4^{\text{th}}$ harmonic the string holds four loops: $L = 4\cdot\dfrac\lambda2 = 80\ \text{m}$.
Solution by Sreeraj P, M.Sc Physics