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.

Answer 3

A stretched wire fixed at both ends supports transverse waves that travel along it with speed \( v = \sqrt{T/\mu} \), where \(T\) is the tension and \(\mu\) is the mass per unit length. For the fundamental (first) mode, the wire's full length \(L\) holds exactly half a wavelength, so \( \lambda = 2L \), and the frequency is \( f = v/\lambda \). With \( T = 4\,\text{N} \) and \( \mu = 0.0001\,\text{kg/m} \), \( v = \sqrt{4/0.0001} = \sqrt{40000} = 200\,\text{m/s} \), and with \( L = 0.5\,\text{m} \), \( \lambda = 1\,\text{m} \), giving \( f = 200/1 = 200\,\text{Hz} \). Let's check the options against this.

1. 100 Hz: This would need either half the wave speed (140 m/s) or twice the wavelength for the same speed; neither matches the given tension and mass per unit length, so it is inconsistent.
2. 200 Hz: Dividing the computed wave speed of 200 m/s by the fundamental wavelength of 1 m gives exactly this value, matching the calculation above.
3. 300 Hz: This would require a wave speed of 300 m/s for the same 1 m wavelength, which does not match \( \sqrt{T/\mu} = 200\,\text{m/s} \) from the given values.
4. 400 Hz: This is exactly double the correct value; it would correspond to the second harmonic (first overtone) of this wire rather than the fundamental.

The wave speed set by the tension and mass per unit length, divided by the fundamental wavelength of twice the wire's length, gives 200 Hz.

So the correct answer is 200 Hz.