Question

A wire $PQ$ has length $4.8\text{ m}$ and mass $0.06\text{ kg}$. Another wire $QR$ has length $2.56\text{ m}$ and mass $0.2\text{ kg}$. Both wires have same radii and are joined as a single wire. This wire is under tension of $80\text{ N}$. A wave pulse of amplitude $3.5\text{ cm}$ is sent along the wire $PQ$ from end $P$. The time taken by the wave pulse to travel along the wire from point $P$ to $R$ is?

• A. $0.1\text{ s}$
• B. $0.12\text{ s}$
• C. $0.14\text{ s}$
• D. $0.16\text{ s}$

Hint:

Since tension $T$ is the same, speed $v \propto 1/\sqrt{\mu}$. Calculate the travel time for each section independently and then sum them up.

Answer 1

The Correct Option is C

Step 1: Calculate the linear mass density ($\mu_{PQ}$) for wire $PQ$: \[ \mu_{PQ} = \frac{m_{PQ{L_{PQ = \frac{0.06\text{ kg{4.8\text{ m = 0.0125\text{ kg/m} \]
Step 2: Calculate the wave speed in wire $PQ$ ($v_{PQ}$) using the formula $v = \sqrt{T/\mu}$: \[ v_{PQ} = \sqrt{\frac{80\text{ N{0.0125\text{ kg/m} = \sqrt{6400} = 80\text{ m/s} \]
Step 3: Calculate the time taken to travel through segment $PQ$ ($t_{PQ}$): \[ t_{PQ} = \frac{L_{PQ{v_{PQ = \frac{4.8\text{ m{80\text{ m/s = 0.06\text{ s} \]
Step 4: Calculate the linear mass density ($\mu_{QR}$) for wire $QR$: \[ \mu_{QR} = \frac{m_{QR{L_{QR = \frac{0.2\text{ kg{2.56\text{ m = 0.078125\text{ kg/m} \]
Step 5: Calculate the wave speed in wire $QR$ ($v_{QR}$): \[ v_{QR} = \sqrt{\frac{80\text{ N{0.078125\text{ kg/m} = \sqrt{1024} = 32\text{ m/s} \]
Step 6: Calculate the time taken to travel through segment $QR$ ($t_{QR}$): \[ t_{QR} = \frac{L_{QR{v_{QR = \frac{2.56\text{ m{32\text{ m/s = 0.08\text{ s} \]
Step 7: Calculate the total time ($t_{total}$) to travel from $P$ to $R$: \[ t_{total} = t_{PQ} + t_{QR} = 0.06\text{ s} + 0.08\text{ s} = 0.14\text{ s} \]