One end – of a thin long uniform string having linear mass density $λ$ is fixed to a wall and other end with a block which can move along a vertical rod. The string is horizontal and has a tension $T_{0}$ . If block executes SHM along the rod with vertical small amplitude A and angular frequency $ω$ . The energy transferred to string in one oscillation of block is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$ω \sqrt{3 T_{0} λ} A^{2} ω π$

$\sqrt{2 T_{0} λ} A^{2} ω π$

$\sqrt{T_{0} λ} A^{2} ω π$

$\sqrt{5 T_{0} λ} A^{2} ω π$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$y = A \sin (ω t - K x) \tan θ = \frac{\partial y}{\partial x} = - A k \cos (ω t - K x)$
At $x = 0; \tan θ = - A K \cos ω t$
Velocity of block $= \frac{\partial y}{\partial t} = A ω \cos (ω t - K x)$
Power supplied $P = (T_{0} \sin θ) (V_{y})$
$= T_{0} [A K \cos ω t] \times [A ω \cos ω t] P = T_{0} A^{2} ω K \cos^{2} ω t = \frac{T_{0} A^{2} ω^{2}}{V} \cos^{2} ω t$
Energy supplied in one oscillation is
$E = \int_{0}^{2 π / ω} P d t = \frac{T_{0} A^{2} ω^{2}}{V} \int_{0}^{2 π / ω} \cos^{2} ω t d t E = \frac{T_{0} A^{2} ω^{2} π}{V} = \sqrt{T_{0} λ} (A^{2} ω π)$