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Two long strings $A$ and $B$, each having linear mass density \[1.2 \times {10^{ - 2}}\;{\rm{kg}}\;{{\rm{m}}^{ - 1}}\], are stretched by different tensions $4.8\,N$ and $7.5\,N$ respectively and are kept parallel to each other with their left ends at x = 0. Wave pulses are produced on the strings at the left ends at $t = 0$ on string $A$ and at $t = 20\,ms$ on string $B$. When and where will the pulse on $B$ overtake that on $A$ ?

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Last updated date: 22nd Jul 2024
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Answer
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Hint: Tension is defined in physics as the pulling force conveyed axially by a thread, cable, cord, or other one-dimensional continuous component, or by each end of a pole, truss member, or other three-dimensional object; tension may also be defined as the action-reaction pair of forces acting at each end of such elements. Tension can be thought of as the polar opposite of compression.

Complete step by step answer:
A pulse is a term used in physics to describe a single disturbance moving through a propagation medium. This medium could be a vacuum (in the case of electromagnetic radiation) or matter, and it could be infinite or finite in size.

Given, \[{{\rm{m}}_{\rm{A}}} = 1.2 \times {10^{ - 2}}\;{\rm{kg}}/{\rm{m}}\]
\[{{\bf{T}}_{\bf{A}}} = {\bf{4}}.{\bf{8N}} \\ \]
\[ \Rightarrow {{\bf{V}}_{\rm{A}}} = \sqrt {{\rm{T}}/{\rm{m}}} = {\bf{20m}}/{\rm{s}} \\ \]
\[\Rightarrow {{\rm{m}}_{\rm{B}}} = 1.2 \times {10^{ - 2}}\;{\rm{kg}}/{\rm{m}} \\\]
\[\Rightarrow {{\rm{T}}_{\rm{B}}} = 7.5\;{\rm{N}} \\\]
\[ \Rightarrow {{\rm{V}}_{\rm{B}}} = \sqrt {{\rm{T}}/{\rm{m}}} \\
\Rightarrow {{\rm{V}}_{\rm{B}}}= {\bf{25\,m}}/{\rm{s}}\]
On string $A$ at $t = 0$
\[{{\rm{t}}_1} = 0 + 20\;{\rm{ms}} = 20 \times {10^{ - 3}} = 0.02\,{\rm{sec}}\]
In 0.02 sec, $A$ has travelled 20 × 0.02 = 0.4m = s
Relative speed between $A$ and $B$ =25−20 = 5m/s = v
Time taken for B to overtake A:
\[\text{Time} = \dfrac{s}{v}\]
\[\Rightarrow \text{Time}= \dfrac{{0.4}}{5}\]
\[\therefore \text{Time}= 0.8\,sec\]

Note: When atoms or molecules are pulled apart and gain potential energy while a restoring force is still present, the restoring force will cause stress. In order to return the string/rod to its relaxed length, each end of a string or rod under a certain strain could tug on the object it is attached to.