Question

The equation of a wave on a string of linear mass density \(0.02 \text{kg m}^{-1}\) is \(Y = 0.01sin[2π(\frac{t}{0.02}-\frac{x}{0.50})]\) m. The tension in the string is

• A. \(12.50\) N
• B. \(6.25\) N
• C. \(25\) N
• D. \(50\) N

Hint:

Read omega and k from the wave equation, find the speed, then use v = root of T over mu.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Compare the given wave with \(y=A\sin(\omega t-kx)\) to find the speed of the wave. Then use \(v=\sqrt{T/\mu}\).

Step 2: Wave speed
The wave is \(y=0.01\sin\left[2\pi\left(\frac{t}{0.02}-\frac{x}{0.50}\right)\right]\). So the period is \(0.02\) s and the wavelength is \(0.50\) m.
\[ v=\frac\lambda T=\frac{0.50}{0.02}=25\ \text{m/s} \]

Step 3: Tension
\[ T=\mu v^2=0.02\times625=12.5\ \text{N} \]

Step 4: Check the options
6.25 N would come from using \(v=12.5\). 25 N and 50 N come from dropping the square. The tension is 12.50 N, option (A).

Final Answer:
The wave speed is 25 m/s, so the tension is 0.02 times 625 = 12.5 N, option (A). \[ \boxed{12.50\ \text{N}} \]