Two small bodies of mass 2 , kg each are atta | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. Initial state: The total mass is $M = 2 + 2 = 4 \, kg$. The spring extension at equilibrium is $x_0 = \frac{Mg}{k} = \frac{4 	imes 10}{200}

Vedclass · Accepted Answer

$70 \, cm$

Vedclass · Answer

(B) $1$. Initial state: The total mass is $M = 2 + 2 = 4 \, kg$. The spring extension at equilibrium is $x_0 = \frac{Mg}{k} = \frac{4 	imes 10}{200} = 0.2 \, m = 20 \, cm$.$2$. After burning the thread: The lower mass falls freely. The upper mass oscillates about a new equilibrium position. The new equilibrium position is $x_{new} = \frac{mg}{k} = \frac{2 	imes 10}{200} = 0.1 \, m = 10 \, cm$ below the natural length of the spring.$3$. The amplitude of oscillation for the upper mass is $A = x_0 - x_{new} = 20 \, cm - 10 \, cm = 10 \, cm$.$4$. The upper mass starts from the original equilibrium position ($20 \, cm$ below natural length) and moves to its highest position (upper extreme),which is $10 \, cm$ above the new equilibrium position ($10 \, cm$ below natural length). Thus,it travels $10 \, cm$ to reach the new equilibrium and another $10 \, cm$ to reach the upper extreme.$5$. Time taken to reach the upper extreme is $t = \frac{T}{2} = \frac{1}{2} 	imes 2\pi \sqrt{\frac{m}{k}} = \pi \sqrt{\frac{2}{200}} = \frac{\pi}{10} \, s$.$6$. In this time,the lower body falls a distance $h = \frac{1}{2}gt^2 = \frac{1}{2} 	imes 10 	imes (\frac{\pi}{10})^2 = 5 	imes \frac{10}{100} = 0.5 \, m = 50 \, cm$.$7$. The upper body is at $10 \, cm$ below the natural length. The lower body is at $20 \, cm$ (original equilibrium) $+ 10 \, cm$ (thread length) $+ 50 \, cm$ (free fall) $= 80 \, cm$ below the natural length.$8$. The distance between them is $80 \, cm - 10 \, cm = 70 \, cm$.

Similar Questions

To make the frequency of a spring oscillator double,we have to

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

When a mass $m$ is attached to a spring,it normally extends by $0.2\, m$. If the mass $m$ is given a slight additional extension and released,what will be its time period?

$A$ block of mass $M$ hangs from a spring and oscillates vertically with an angular frequency $\omega$. If the block is removed from the spring,when it is in equilibrium position,the spring shortens by

Vedclass Test Series

Exam Paper Generator

Online Exam Module