Question

A wire 0.5 m long and with a mass per unit length of 0.0001 kg/m vibrates under a tension of 4 N. The fundamental frequency is:

• A. 100 Hz
• B. 200 Hz
• C. 300 Hz
• D. 400 Hz

Hint:

The fundamental frequency of a string depends on its length, tension, and mass per unit length.

Answer 1

The Correct Option is B

To find the fundamental frequency of a vibrating wire, we use the formula for the fundamental frequency of a string fixed at both ends:

f = \(\frac{1}{2L}\) \(\sqrt{\frac{T}{\mu}}\)

Where:

• f is the fundamental frequency. 
• L is the length of the wire (0.5 m).
• T is the tension in the wire (4 N).
• \(\mu\) is the mass per unit length of the wire (0.0001 kg/m).

Substitute these values into the formula:

f = \(\frac{1}{2 \times 0.5}\) \(\sqrt{\frac{4}{0.0001}}\)

f = \(\frac{1}{1}\) \(\sqrt{40000}\)

The square root of 40000 is 200, so:

f = 200 Hz

Hence, the fundamental frequency of the wire is 200 Hz.

Answer 2

The fundamental frequency \( f_1 \) of a vibrating string is given by: \[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where: 
- \( L = 0.5 \, \text{m} \),
- \( T = 4 \, \text{N} \),
- \( \mu = 0.0001 \, \text{kg/m} \). Substitute the values into the equation: \[ f_1 = \frac{1}{2 \times 0.5} \sqrt{\frac{4}{0.0001}} = \frac{1}{1} \times \sqrt{40000} = 200 \, \text{Hz} \] Thus, the fundamental frequency is 200 Hz.