Q. A string is hanging from a rigid support. A transverse pulse is excited at its free end. The speed at which the pulse travels a distance $x$ is proportional to
A) $x$
B) $\frac{1}{x}$
C) $\frac{1}{\sqrt{x}}$
D) $\sqrt{x}$
Explanation
(d) Let mass and length of the string are $M\ $and $L$
Pulse at distance $x$ from free end
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Mass per unit length $= M/L$
$\therefore$ Mass of length $x$ i.e.
Tension $T = \left( \frac{M}{L} \right)xg$
$\therefore$ Velocity $= \sqrt{T/m}$
$\Rightarrow v = \sqrt{\frac{Mgx/L}{M/L}} = \sqrt{gx}$
$\Rightarrow v \propto \sqrt{x}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.