Question:
A string is clamped at both the ends and it is vibrating in its 4th 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.)
A string is clamped at both the ends and it is vibrating in its 4th 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.)
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Compare with \(Y = A\sin(kx)\cos(\omega t)\) to get \(k\); a string fixed at both ends in the n-th harmonic has \(kL = n\pi\).
Updated On: Nov 5, 2026
- \(60\) m
- \(20\) m
- \(80\) m
- \(40\) m
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The Correct Option is C
Solution and Explanation
Comparing with \(Y = A\sin kx\cos\omega t\): \(k = 0.157\ \mathrm{m^{-1}}\), so \(\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{0.157} = 40\) m.
For the 4th harmonic, \(L = 4\cdot\frac{\lambda}{2} = 2\lambda = 80\) m.
Answer: (3) \(80\) m
For the 4th harmonic, \(L = 4\cdot\frac{\lambda}{2} = 2\lambda = 80\) m.
Answer: (3) \(80\) m
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