Question

The equation of a transverse wave propagating on a stretched string is given by $y = 3\sin(4x + 200t)$, where $x$ and $y$ are in metre and the time $t$ is in second. If the tension applied to the string is $500\,\mathrm{N}$, the linear density of the string is

• A. $0.25\,\mathrm{kg\,m^{-1}}$
• B. $0.4\,\mathrm{kg\,m^{-1}}$
• C. $0.2\,\mathrm{kg\,m^{-1}}$
• D. $0.1\,\mathrm{kg\,m^{-1}}$

Hint:

Speed of a transverse wave on a string depends on Tension ($T$) and mass per unit length ($\mu$): $v = \sqrt{T/\mu}$.

Answer 1

The Correct Option is C

Step 1: Extract Wave Parameters:
Comparing $y = 3\sin(4x + 200t)$ with standard equation $y = A\sin(kx + \omega t)$: Wave number $k = 4\,\mathrm{m^{-1}}$. Angular frequency $\omega = 200\,\mathrm{rad/s}$.
Step 2: Calculate Wave Velocity ($v$):
\[ v = \frac{\omega}{k} = \frac{200}{4} = 50\,\mathrm{m/s} \]
Step 3: Relate Velocity to Tension and Linear Density:
For a stretched string, $v = \sqrt{\frac{T}{\mu}}$, where $T$ is tension and $\mu$ is linear mass density. Squaring both sides: $v^2 = \frac{T}{\mu} \implies \mu = \frac{T}{v^2}$.
Step 4: Calculation:
Given $T = 500\,\mathrm{N}$. \[ \mu = \frac{500}{(50)^2} = \frac{500}{2500} = \frac{1}{5} = 0.2\,\mathrm{kg\,m^{-1}} \]