Question

Two masses ${m_1} = 1kg$and ${m_2} = 0.5kg$ are suspended together by a massless spring of spring constant $12.5N/m$ when masses are in equilibrium ${m_1}$ is removed without disturbing the system New amplitude of oscillation will be
a. 30cm
b. 50cm
c. 80cm
d. 60cm

Answer 1

Hint Before the equilibrium its length is same as before and after the equilibrium and the mass is removed then the new length is gained and the g here will be 10m/s
Formula Used
Below formula will be used to solve above numerical:
$\dfrac{{{m_1}g}}{k} = \Delta l$
Where g is the gravitational force
 ∆l is the difference of extended length to the old length
K is the spring constant
M is the masses given here

Complete Step by step by step solution
As we know that the amplitude is the magnitude of change in the oscillating variable with each oscillation within an oscillating system.
Also the block begins to oscillate in SHM between \[x = + A{{ }}and{{ }}x = - A,\]where A is the amplitude of the motion and T is the period of one oscillation
So as per the question requirement or saying,
Let us assume initially that ${m_1} + {m_2}$ are attached to string and extension of string by this body is $l$
We can obtain here an equation that $({m_2})g = Kl$$({m_1} + {m_2})g = Kl$……………..(1)
So here the spring constant will be (k) $12.5N/m$
When masses are in equilibrium ${m_1}$ is removed without disturbing the system so the length attained is new length that is $l'$
 We can obtain here an new equation that $({m_2})g = Kl$…………(2)
So by combining the both of the equations (1) and (2)
${m_1}g + Kl' = Kl$
\[{m_1}g = K(l' - l)\]
${m_1}g = K\Delta l$
$\dfrac{{{m_1}g}}{k} = \Delta l$
$\dfrac{{1 \times 10}}{{12.5}} = \Delta l$$ = 0.8m = 80cm$

Hence option C is correct.

Note
Key point to be remembered is that the spring constant does not change or vary.
 When it attains an equilibrium it has different length and before it is different length and also when masses are removed it acquires new lengths.