Question

The string of length 2 m is fixed at both ends. If the string vibrates in its fourth normal mode with a frequency of 500 Hz, then the waves would travel on it with a velocity of

• A. \(125 \, \text{m/s}\)
• B. \(250 \, \text{m/s}\)
• C. \(500 \, \text{m/s}\)
• D. \(1000 \, \text{m/s}\)

Hint:

In standing wave problems on strings, remember that the \(n^{\text{th}}\) mode has \(n\) loops and the wavelength is \( \frac{2L}{n} \). Use \(v = f \lambda\) to find the wave speed.

Answer 1

The Correct Option is C

When a string is fixed at both ends and vibrates in its \(n^{\text{th}}\) normal mode, the wavelength of the standing wave is given by: \[ \lambda = \frac{2L}{n} \] Given: - \(L = 2 \, \text{m}\) (length of the string)
- \(n = 4\) (fourth normal mode)
- \(f = 500 \, \text{Hz}\) (frequency)
Using the formula for wavelength: \[ \lambda = \frac{2 \times 2}{4} = \frac{4}{4} = 1 \, \text{m} \] Now, using the wave velocity formula: \[ v = f \lambda = 500 \times 1 = 500 \, \text{m/s} \] Thus, the wave travels with a velocity of \(500 \, \text{m/s}\).