Question

A transverse displacement of vibrating string is \( y = 0.06 \sin\left(\frac{2 \pi}{3} x\right) \cos(120 \pi t) \). If the mass per unit length of a string is \( 4 \times 10^{-2} \) kg/m, then the tension in the string will be

• A. 1296 N
• B. 1000 N
• C. 720 N
• D. 500 N

Hint:

Compare the given standing wave equation with the standard form to get the wave speed, then use the speed formula for a string.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
A standing wave has the form \(y = A\sin(kx)\cos(\omega t)\), where the wave speed is \(v = \dfrac{\omega}{k}\).

Step 2: Find k and Wave Speed
Here \(k = \dfrac{2\pi}{3}\) and \(\omega = 120\pi\). So \(v = \dfrac{120\pi}{2\pi/3} = 180\) m/s.

Step 3: Tension
Since \(v = \sqrt{\dfrac{T}{\mu}}\), \(T = \mu v^2 = 4 \times 10^{-2} \times (180)^2 = 4 \times 10^{-2} \times 32400 = 1296\) N.

Step 4: Compare the options
The tension is 1296 N, the first option.

Final Answer:
The tension in the string is 1296 N. \[ \boxed{\text{(A) }1296\ \text{N}} \]