Question

The equation of vibration of a stretched string fixed at both ends and vibrating in 5th harmonic is \( Y = 3 \sin(0.4x) \cos(200\pi t) \), where \( x \) and \( Y \) are in cm and \( t \) in seconds. The length of the string is

• A. \( (10.5) \pi \, \text{cm} \)
• B. \( (8.5) \pi \, \text{cm} \)
• C. \( (12.5) \pi \, \text{cm} \)
• D. \( (4.5) \pi \, \text{cm} \)

Hint:

In vibrating strings, the wavelength and length of the string are related by the harmonic number. The string length for the nth harmonic can be found by the formula \( L = \frac{n \pi}{k} \).

Answer 1

The Correct Option is C

Step 1: Understanding the equation of vibration.
The general equation for the vibration of a string in the nth harmonic is: \[ Y = A \sin \left( \frac{n \pi x}{L} \right) \cos (\omega t) \] Where: - \( n = 5 \) (since it's the 5th harmonic), - \( A = 3 \) (amplitude), - \( \omega = 200\pi \), - \( L \) is the length of the string. Step 2: Relating the equation to the given information.
Comparing the given equation \( Y = 3 \sin(0.4x) \cos(200\pi t) \) with the standard form: \[ \frac{n \pi}{L} = 0.4 \] Step 3: Solving for \( L \).
Solving for the length of the string: \[ L = \frac{n \pi}{0.4} = \frac{5 \pi}{0.4} = 12.5 \pi \, \text{cm} \] Thus, the length of the string is \( 12.5 \pi \, \text{cm} \), corresponding to option (C).