Question

A transverse wave is propagating on the string. The linear density of vibrating string is \(10^{-3}\) kg/m. The equation of the wave is \(Y=0.05\,\sin (x+15t)\) where \(x\) and \(Y\) are in metre and time in second. The tension in the string is

• A. \(0.2\ \text{N}\)
• B. \(0.25\ \text{N}\)
• C. \(0.225\ \text{N}\)
• D. \(0.325\ \text{N}\)

Hint:

Get the wave speed from the coefficients of x and t, then use the speed formula for a string to find tension.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The wave speed on a string is \(v=\sqrt{\frac{T}{\mu}}\), so \(T=\mu v^{2}\).

Step 2: Wave Speed
Comparing \(Y=0.05\sin(x+15t)\) with \(\sin(kx+\omega t)\): \(k=1\), \(\omega=15\).
\(v=\frac{\omega}{k}=15\ \text{m/s}\).

Step 3: Calculation
\(T=10^{-3}\times15^{2}=10^{-3}\times225=0.225\ \text{N}\).

Final Answer:
The tension in the string is 0.225 N. \[ \boxed{\text{(C) }0.225\ \text{N}} \]