Question 355: Waves

Question

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

Accepted Answer

To determine the time taken by the wave pulse to travel along the wire from point P to R, we first need to calculate the speed of the wave in each segment of the wire (PQ and QR) and then find the time taken to travel through each segment.
We have two wires PQ and QR with different lengths and masses but the same radius and therefore the same cross-sectional area A. Both wires are under the same tension T. Given that the tension and cross-sectional area are the same for both wires, we can assume that the linear density (mass per unit length) is different because they have different masses and lengths.
To find the linear density (ÃÂ¼) for each wire, we use the formula:
$$ \mu = \frac{m}{L} $$
where
m is the mass of the wire, and
L is the length of the wire.
For wire PQ:
$$ \mu_{PQ} = \frac{m_{PQ}}{L_{PQ}} = \frac{0.06\ kg}{4.8\ m} = 0.0125\ kg/m $$
For wire QR:
$$ \mu_{QR} = \frac{m_{QR}}{L_{QR}} = \frac{0.2\ kg}{2.56\ m} = 0.078125\ kg/m $$
The speed of a wave in a stretched string or wire is given by the formula:
$$ v = \sqrt{\frac{T}{\mu}} $$
where
v is the speed of the wave, and
T is the tension in the wire.
Using this formula, we can calculate the speed of the wave in each section of the wire.
For wire PQ:
$$ v_{PQ} = \sqrt{\frac{T}{\mu_{PQ}}} = \sqrt{\frac{80\ N}{0.0125\ kg/m}} = \sqrt{6400\ m^2/s^2} = 80\ m/s $$
For wire QR:
$$ v_{QR} = \sqrt{\frac{T}{\mu_{QR}}} = \sqrt{\frac{80\ N}{0.078125\ kg/m}} = \sqrt{1024\ m^2/s^2} = 32\ m/s $$
Now we will find the time taken for the wave to travel through each section.
Time is distance over speed, so for wire PQ:
$$ t_{PQ} = \frac{L_{PQ}}{v_{PQ}} = \frac{4.8\ m}{80\ m/s} = 0.06\ s $$
And for wire QR:
$$ t_{QR} = \frac{L_{QR}}{v_{QR}} = \frac{2.56\ m}{32\ m/s} = 0.08\ s $$
To find the total time taken by the wave pulse to travel from P to R, we add the times for PQ and QR:
$$ t_{total} = t_{PQ} + t_{QR} = 0.06\ s + 0.08\ s = 0.14\ s $$
The correct answer is:
Option C: 0.14 s

Answer

0.1 s

Answer

0.12 s

Answer

0.14 s

Answer

0.16 s

Waves

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