Question

If the speed of the transverse wave in a wire under certain tension T is v, then its speed under tension $2T$ (in $ms^{-1}$) is

• A. $\frac{v}{\sqrt{2}}$
• B. $2v$
• C. $\sqrt{2}v$
• D. $\frac{3v}{\sqrt{2}}$
• E. $\frac{v}{2}$

Hint:

$v \propto \sqrt{T}$; doubling tension increases speed by a factor of $\sqrt{2} \approx 1.41$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem deals with the properties of transverse waves on a stretched string or wire. The speed of such a wave depends on the tension in the wire and its linear mass density.
Step 2: Key Formula or Approach:
The speed (v) of a transverse wave on a wire is given by the formula: \[ v = \sqrt{\frac{T}{\mu}} \] where T is the tension in the wire and \( \mu \) is the linear mass density (mass per unit length) of the wire.
From this formula, we can see that if the linear mass density \( \mu \) is constant, the speed is directly proportional to the square root of the tension: \[ v \propto \sqrt{T} \] This allows us to set up a ratio for two different tensions. \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \] Step 3: Detailed Explanation:
We are given two situations:
Situation 1: - Tension, \( T_1 = T \) - Speed, \( v_1 = v \) Situation 2: - Tension, \( T_2 = 2T \) - Speed, \( v_2 = ? \) Using the ratio from Step 2: \[ \frac{v_2}{v} = \sqrt{\frac{2T}{T}} \] Cancel the common factor T: \[ \frac{v_2}{v} = \sqrt{2} \] Now, solve for \( v_2 \): \[ v_2 = \sqrt{2} v \] Step 4: Final Answer:
The new speed of the wave is \( \sqrt{2}v \).