Question

If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

• A. 1:1
• B. √2:1
• C. 1:√2
• D. 1:2

Answer 1

The Correct Option is C

To solve this problem, we need to understand the relationship between tension and the speed of a transverse wave on a string. The speed \( v \) of a wave on a string is given by the equation: \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string. 

If the initial tension is \( T_1 \) and the initial speed of the wave is \( v_1 = \sqrt{\frac{T_1}{\mu}} \), then when the tension is doubled, the new tension \( T_2 \) is \( 2T_1 \). The new speed \( v_2 \) is \( v_2 = \sqrt{\frac{2T_1}{\mu}} \).

To find the ratio of the initial and final speeds, we calculate:

\( \frac{v_1}{v_2} = \frac{\sqrt{\frac{T_1}{\mu}}}{\sqrt{\frac{2T_1}{\mu}}} = \frac{\sqrt{T_1}}{\sqrt{2T_1}} = \frac{1}{\sqrt{2}} \).

Thus, the ratio of the initial speed to the final speed is \( 1:\sqrt{2} \).

Initial Speed \(v_1\)	\(\sqrt{\frac{T_1}{\mu}}\)
Final Speed \(v_2\)	\(\sqrt{\frac{2T_1}{\mu}}\)
Ratio \(v_1:v_2\)	1:\(\sqrt{2}\)

Answer 2

The correct answer is (C) : \(1:\sqrt2\)
Given :
Tension in the string = T,
Mass per unit length = μ,
Velocity of transverse wave on a string , v = \(\sqrt{\frac{\bar{T}}{\mu}}\)
\(v_1=\sqrt{\frac{T_1}{\mu}}\)
\(v_2=\sqrt{\frac{T_2}{\mu}}\)
As per question T₂ = 2T₁
\(\frac{v_1}{v_2}=\sqrt{\frac{\bar{1}}{2}}=\frac{1}{\sqrt2}\)