Question

Two mechanical waves on strings of equal length (\( L \)) and tension (\( T \)) having linear mass density \( \mu_1/\mu_2 = 1/2 \). Find the ratio of time taken for a wave pulse to travel from one end to the other in both strings. (Ignore gravity)

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \sqrt{2} \)
• D. 2

Hint:

The velocity of a wave on a string depends on the tension and mass density of the string. For strings of equal length and tension, the ratio of the times taken is the inverse square root of the ratio of the mass densities.

Answer 1

The Correct Option is B

Step 1: Use the formula for wave velocity on a string.
The velocity \( v \) of a wave on a string is given by: \[ v = \sqrt{\frac{T}{\mu}} \] Where \( T \) is the tension and \( \mu \) is the linear mass density. The time taken for the wave pulse to travel the length of the string is: \[ t = \frac{L}{v} \] Step 2: Compare the time for both strings.
For the first string: \[ t_1 = \frac{L}{\sqrt{T/\mu_1}} = L \sqrt{\frac{\mu_1}{T}} \] For the second string: \[ t_2 = \frac{L}{\sqrt{T/\mu_2}} = L \sqrt{\frac{\mu_2}{T}} \] Step 3: Find the ratio of times.
The ratio of the times is: \[ \frac{t_1}{t_2} = \sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \] Step 4: Conclusion.
The ratio of times taken is \( \frac{1}{\sqrt{2}} \), which corresponds to option (2).