Question

A stretched string under tension fixed at both ends vibrates in the 4th harmonic. The equation of the stationary wave is \( Y = 3 \sin(200\pi t) \cos(0.4x) \), where \( x \) and \( y \) are in cm and \( t \) in seconds. The length of the vibrating string is

• A. \( 10\pi \)
• B. \( 4\pi \)
• C. \( 6\pi \)
• D. \( 8\pi \)

Hint:

The length of the vibrating string for the \( n \)-th harmonic is \( L = \frac{n\lambda}{2} \), where \( \lambda \) is the wavelength of the wave.

Answer 1

The Correct Option is A

Step 1: Understand the form of the wave equation.
The equation for a stationary wave on a string is given by: \[ Y(x,t) = A \sin(\omega t) \cos(kx) \] where: - \( \omega \) is the angular frequency, - \( k \) is the wave number. Comparing the given equation \( Y = 3 \sin(200\pi t) \cos(0.4x) \) with the standard form, we find: \[ \omega = 200\pi \quad \text{and} \quad k = 0.4 \] Step 2: Find the wavelength.
The wavelength \( \lambda \) is related to the wave number \( k \) by: \[ k = \frac{2\pi}{\lambda} \] Substituting \( k = 0.4 \): \[ 0.4 = \frac{2\pi}{\lambda} \quad \Rightarrow \quad \lambda = \frac{2\pi}{0.4} = 5\pi \] Step 3: Use the harmonic number.
For the 4th harmonic, the length of the vibrating string \( L \) is given by: \[ L = \frac{n\lambda}{2} \] where \( n \) is the harmonic number. Substituting \( n = 4 \) and \( \lambda = 5\pi \): \[ L = \frac{4 \times 5\pi}{2} = 10\pi \] Step 4: Conclusion.
Thus, the length of the vibrating string is \( 10\pi \), which corresponds to option (A).