Question

The linear mass density of a stretched string is $\mu$ and tension is T. The wave velocity is:

• A. $\sqrt{T/\mu}$
• B. $\sqrt{\mu/T}$
• C. $T/\mu$
• D. $\mu/T$

Hint:

To increase the speed of a wave in a string, you can either increase the tension ($T$) by stretching it tighter, or use a lighter string (decrease $\mu$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the mathematical expression of the speed of a transverse wave traveling along a stretched string.
This is a standard relation in wave mechanics.

Step 2: Key Formula or Approach:
The velocity of a transverse wave in a stretched string depends on the tension ($T$) in the string and its linear mass density ($\mu$).
We can verify this formula using dimensional analysis.

Step 3: Detailed Explanation:

• The wave velocity ($v$) is determined by the restoring force (tension, $T$) and the inertial property (linear mass density, $\mu$).

• Let us check the dimensions of each quantity:
1. Tension ($T$) is a force, so its dimensional formula is:
\[ [T] = [M^1 L^1 T^{-2}] \]
2. Linear mass density ($\mu$) is mass per unit length ($m/L$), so its dimensional formula is:
\[ [\mu] = [M^1 L^{-1} T^0] \]

• Let us find the dimensions of the ratio $T/\mu$:
\[ \left[\frac{T}{\mu}\right] = \frac{[M^1 L^1 T^{-2}]}{[M^1 L^{-1}]} = [L^2 T^{-2}] \]

• Taking the square root of this ratio:
\[ \left[\sqrt{\frac{T}{\mu}}\right] = \sqrt{[L^2 T^{-2}]} = [L^1 T^{-1}] \]

• The dimensional formula $[L^1 T^{-1}]$ corresponds exactly to velocity.

• This confirms the physical derivation, which yields the wave velocity as:
\[ v = \sqrt{\frac{T}{\mu}} \]


Step 4: Final Answer:
The velocity of the wave in a stretched string is given by the expression $\sqrt{T/\mu}$.