Question

A horizontal stretched elastic string has a length 3m and mass 12kg. It is subjected to a tension of 1.6 N. Transverse waves of 40Hz are propagated down the string. Calculate the distance between successive crests of the motion.

Answer 1

To find the distance between successive crests of the transverse waves (the wavelength), we can use the wave speed formula and the relationship between frequency, wavelength, and wave speed.

1. **Calculate the wave speed (v)**:
The wave speed on a stretched string is given by the formula:

\[
v = \sqrt{\frac{T}{\mu}}
\]

where:
- \( T \) is the tension in the string.
- \( \mu \) is the linear mass density of the string.

The linear mass density \( \mu \) is given by:

\[
\mu = \frac{m}{L}
\]

where:
- \( m = 12 \, \text{kg} \) is the mass of the string.
- \( L = 3 \, \text{m} \) is the length of the string.

So,

\[
\mu = \frac{12 \, \text{kg}}{3 \, \text{m}} = 4 \, \text{kg/m}
\]

Now, substituting \( \mu \) into the wave speed formula:

\[
v = \sqrt{\frac{1.6 \, \text{N}}{4 \, \text{kg/m}}}
\]

Calculating the wave speed:

\[
v = \sqrt{\frac{1.6}{4}} = \sqrt{0.4} \approx 0.632 \, \text{m/s}
\]

2. **Calculate the wavelength (\( \lambda \))**:
The relationship between wave speed, frequency, and wavelength is given by:

\[
v = f \lambda
\]

Rearranging for wavelength gives us:

\[
\lambda = \frac{v}{f}
\]

where \( f = 40 \, \text{Hz} \).

Substituting the values:

\[
\lambda = \frac{0.632 \, \text{m/s}}{40 \, \text{Hz}} = \frac{0.632}{40} \approx 0.0158 \, \text{m}
\]

3. **Final Result**:
The distance between successive crests of the motion, which is the wavelength, is approximately \( 0.0158 \, \text{m} \), or \( 15.8 \, \text{mm} \).

Thus, the distance between successive crests of the motion is approximately **0.0158 meters** or **15.8 millimeters**.