Vibration of Stretched String in Fifth Harmonic

To determine the correct statements, we analyze the given standing wave equation for a string fixed at both ends:

$y(x, t) = (0.01\text{ m}) \sin [(62.8\text{ m}^{-1}) x] \cos [(628\text{ s}^{-1})t]$

Comparing this with the standard equation for a standing wave, $y(x, t) = A \sin(kx) \cos(\omega t)$, we get:

• Amplitude constant, $A = 0.01\text{ m}$
• Wave number, $k = 62.8\text{ m}^{-1}$
• Angular frequency, $\omega = 628\text{ s}^{-1}$
• Given $\pi = 3.14$

1. Calculation of Wavelength ($\lambda$) and Frequency ($f_5$)

The wavelength $\lambda$ is given by: $k = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{62.8\text{ m}^{-1}} = \frac{6.28}{62.8}\text{ m} = 0.1\text{ m}$

The frequency of vibration in the $5^{\text{th}}$ harmonic ($f_5$) is: $\omega = 2\pi f_5 \implies f_5 = \frac{\omega}{2\pi} = \frac{628\text{ s}^{-1}}{2 \times 3.14} = \frac{628}{6.28}\text{ Hz} = 100\text{ Hz}$

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2. Analysis of Options

• Option (A): Number of Nodes For a string fixed at both ends vibrating in the $n^{\text{th}}$ harmonic, the number of loops (antinodes) is $n$, and the total number of nodes is $n + 1$. Since the string is vibrating in its fifth harmonic ($n = 5$): $\text{Total number of nodes} = 5 + 1 = 6$ Therefore, Option (A) is incorrect.

• Option (B): Length of the String ($L$) For the $5^{\text{th}}$ harmonic of a string fixed at both ends: $L = 5\left(\frac{\lambda}{2}\right) = 5 \times \left(\frac{0.1\text{ m}}{2}\right) = 5 \times 0.05\text{ m} = 0.25\text{ m}$ Therefore, Option (B) is correct.

• Option (C): Maximum Displacement at the Midpoint The position of the midpoint of the string is: $x_{\text{mid}} = \frac{L}{2} = \frac{0.25\text{ m}}{2} = 0.125\text{ m}$ The spatial phase term at the midpoint is: $k x_{\text{mid}} = (62.8\text{ m}^{-1}) \times (0.125\text{ m}) = \left(\frac{2\pi}{0.1}\right) \times 0.125 = \frac{5\pi}{2}\text{ rad}$ Evaluating the spatial factor: $\left|\sin\left(\frac{5\pi}{2}\right)\right| = \left|\sin\left(2\pi + \frac{\pi}{2}\right)\right| = \sin\left(\frac{\pi}{2}\right) = 1$ Thus, the amplitude of oscillation at the midpoint is: $y_{\text{max}}\left(x_{\text{mid}}\right) = 0.01\text{ m} \times 1 = 0.01\text{ m}$ Therefore, Option (C) is correct.

• Option (D): Fundamental Frequency ($f_1$) The frequency of the $n^{\text{th}}$ harmonic is related to the fundamental frequency $f_1$ by $f_n = n f_1$. For $n = 5$: $f_5 = 5 f_1 \implies f_1 = \frac{f_5}{5} = \frac{100\text{ Hz}}{5} = 20\text{ Hz}$ Therefore, Option (D) is incorrect.

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Conclusion

The correct statements are B and C.