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A wire of length $50 \mathrm{~cm}$ and weighing $10 \mathrm{gm}$ is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of $50 \mathrm{~N} / \mathrm{m}$ and is stretched by $1 \mathrm{~cm}$. If a wave pulse is produced on the string near the wall, then how much time will it take to reach the spring?
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Verified Answer
The correct answer is:
$0.1 \mathrm{~s}$
Tension in wire $=k x=50 \times \frac{1}{100}=0.5 \mathrm{~N}$ Mass per unit length of string $\mu=\frac{m}{l}$
$$
=10 \times 10^{-3}=\frac{1}{50} \mathrm{kgm}^{-1}=50 \times 10^{-2}
$$
Speed of wave pulse on string,
$$
v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{0.5}{\left(\frac{1}{50}\right)}}=\sqrt{25} \mathrm{~ms}^{-1}=5 \mathrm{~ms}^{-1}
$$
Time required by pulse to each other end,
$$
t=\frac{l}{v}=\frac{50 \times 10^{-2} \mathrm{~m}}{5 \mathrm{~ms}^{-1}}=0.1 \mathrm{~s}
$$
$$
=10 \times 10^{-3}=\frac{1}{50} \mathrm{kgm}^{-1}=50 \times 10^{-2}
$$
Speed of wave pulse on string,
$$
v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{0.5}{\left(\frac{1}{50}\right)}}=\sqrt{25} \mathrm{~ms}^{-1}=5 \mathrm{~ms}^{-1}
$$
Time required by pulse to each other end,
$$
t=\frac{l}{v}=\frac{50 \times 10^{-2} \mathrm{~m}}{5 \mathrm{~ms}^{-1}}=0.1 \mathrm{~s}
$$
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