Q. A string of mass 0.2 kg m has length $l = 0.6$m. It I fixed at both ends and stretched such that it has a tension of 80 N. The string vibrates in three segments with amplitude=0.5 cm. The amplitude of transverse velocity is
A) $9.42ms^{- 1}$
B) $3.14ms^{- 1}$
C) $1.57ms^{- 1}$
D) $6.28ms^{- 1}$
Explanation
(c) As the string is vibrating in three segments, therefore,
$l = \frac{3\lambda}{2}\ or\ \lambda = \frac{2l}{3} = \frac{2(0.6)}{3} = 0.4m$
As $v = \sqrt{\frac{T}{m}}\ \ \ \therefore v = \sqrt{\frac{80}{0.2}} = 20ms^{- 1}$
$n = \frac{v}{\lambda} = \frac{20}{0.4} = 50Hz$
Amplitude of particle velocity
$= \left( \frac{dy}{dt} \right)_{\max} = \left( a_{\max} \right)\omega = a_{\max}(2\pi n)$
$\left( 0.5 \times 10^{- 2} \right) \times 2\pi \times 50 = 1.57ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.