Two masses m_1=1 ,kg and m_2=0.5 ,kg are susp | vedclass

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

Question

(c)

Points of equilibrium of the spring will be when no force acts on it.

$k x=\left(m_1+m_2\right) g$

$x=\frac{\left(m_1+m_2\right) g}{k}$

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(c)

Points of equilibrium of the spring will be when no force acts on it.

$k x=\left(m_1+m_2ight) g$

$x=\frac{\left(m_1+m_2ight) g}{k}$

The new equilibrium position which will be the mean position of $S.H.M.$ will be simply $\frac{m_2 g}{k}$

New amplitude will be maximum displacement from $\frac{m_2 g}{k}$ which is :

$A=\frac{\left(m_1+m_2ight) g}{k}-\frac{m_2 g}{k}$

or $A=\frac{m_1 g}{k}$

or $A=\frac{1 	imes 10}{12.5}$

or $A=\frac{4}{5} \,m$

$	herefore A=0.8 \,m$ or $80 \,cm$

Two masses $m_1=1 \,kg$ and $m_2=0.5 \,kg$ are suspended together by a massless spring of spring constant $12.5 \,Nm ^{-1}$. When masses are in equilibrium $m_1$ is removed without disturbing the system. New amplitude of oscillation will be .......... $cm$

Similar Questions

Two springs have spring constants ${K_A}$ and ${K_B}$ and ${K_A} > {K_B}$. The work required to stretch them by same extension will be

A $15 \,g$ ball is shot from a spring gun whose spring has a force constant of $600 \,N/m$. The spring is compressed by $5 \,cm$. The greatest possible horizontal range of the ball for this compression is .... $m$ ($g = 10 \,m/s^2$)

In the situation as shown in figure time period of vertical oscillation of block for small displacements will be

A mass $m$ is attached to two springs of same force constant $K$, as shown in following four arrangements. If $T_1, T_2, T_3$ and $T_4$ respectively be the time periods of oscillation in the following arrangements, in which case time period is maximum?

If a vertical mass spring system is taken to the moon, will its time period after ?