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Physics Question on Oscillations

A mass MM is suspended from a light spring. An additional mass mm added displaces the spring further by a distance XX. Now the combined mass will oscillate on the spring with period

A

T=2πmgX(M+m)T= 2\pi \sqrt{\frac{mg}{X\left(M+m\right)}}

B

T=2πX(M+m)mgT= 2\pi \sqrt{\frac{X\left(M+m\right)}{mg}}

C

T=π2X(M+m)mgT= \frac{\pi}{2} \sqrt{\frac{X\left(M+m\right)}{mg}}

D

T=2π(M+m)mgT= 2\pi \sqrt{\frac{\left(M+m\right)}{mg}}

Answer

T=2πX(M+m)mgT= 2\pi \sqrt{\frac{X\left(M+m\right)}{mg}}

Explanation

Solution

Time period of spring block system is given by,
T=2πmass of blockspring constant{T= 2\pi \sqrt{\frac{mass \,of \, block }{spring \, constant}}}
Here, mass of block = (M+m)(M + m)
Spring constant k=mgXk = \frac{mg}{X}
[∵  At equilibrium,(M+m)g=k(X0+X) or,mg=kX(Initially,Mg=kX0)][ \because \:\: {At \, equilibrium}, (M +m)g = k (X_0 + X ) \, {or} , mg = kX ( {Initially}, Mg = kX_0)]
∴   T=2π(M+m)mgX=2π(M+m)Xmg\therefore \:\:\: T= 2\pi \sqrt{\frac{\left(M +m\right)}{\frac{mg}{X}}} = 2\pi \sqrt{\frac{\left(M + m\right)X}{mg}}