A 4 kg block is suspended from the ceiling of an elevator through a spring having a linear mass density of 19.2 × 10^ -3 ~kg ~m^ -3 . Find the speed with respect to spring with which a wave pulse can proceed on the

Correct answer: D. 50 m/s Given, m=19.2 × 10^ -3 ~kg / m Now see the free body diagram. The equation for the motion aligned T-4 g-4 a & =0 T & =4(a+g)=4(2+10) & =48 ~N aligned Now wave speed, aligned v & = ((T)/(m)) & = 2.5 × 10^ -3 =50 ~m / s aligned

JIPMER 2015

Exam: JIPMER
Year: 2015
Subject: Physics
Chapter: Waves and Sound

JIPMER 2015 · Physics · Waves and Sound

A 4 kg block is suspended from the ceiling of an elevator through a...

A 4 kg block is suspended from the ceiling of an elevator through a spring having a linear mass density of

19.2 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-3}

. Find the speed with respect to spring with which a wave pulse can proceed on the spring if the elevator accelerates up at the rate of

2.0 \mathrm{~ms}^{-2}

. Take

g=10 \mathrm{~ms}^{-2}

A 4 kg block is suspended from the ceiling of an elevator through a spring having a linear mass density of 19.2 × 10^

Options

A30 m/s
B42 m/s
C46 m/s
D50 m/s

Correct answer

D. 50 m/s

Step-by-step solution

Given,

m=19.2 \times 10^{-3} \mathrm{~kg} / \mathrm{m}

Now see the free body diagram. The equation for the motion

\begin{aligned} T-4 g-4 a & =0 \\ T & =4(a+g)=4(2+10) \\ & =48 \mathrm{~N}\end{aligned}

Now wave speed,

\begin{aligned} v & =\sqrt{\left(\frac{T}{m}\right)} \\ & =\sqrt{2.5 \times 10^{-3}}=50 \mathrm{~m} / \mathrm{s}\end{aligned}

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