First slide

Speed of wave

Question

A string of uniform cross-sectional area is suspended from one of its ends. A transverse pulse is produced at its lower end and the pulse starts propagating up the string. Then the acceleration of the pulse is (Take g = 10 m/s²)

Difficult

A

5 m/s²
B

10 m/s²
C

zero
D

8 m/s²

Solution

At a distance x from the lower end, tension in the string $T_{x} = μ . x . g$ where $μ$ is the mass per unit length of the string.
$∴ V_{x} = \sqrt{\frac{T_{x}}{μ}} = \sqrt{gx}$
$∴$ Acceleration $a = V_{x} . \frac{{dV}_{x}}{dx} = \sqrt{gx} \frac{1}{2} \sqrt{\frac{g}{x}} = \frac{g}{2} = 5 m / s^{2}$

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