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.