JEE Main 2021 · 24 Feb · Shift 1 · Physics Q11

← JEE Main 2021 · 24 Feb · Shift 1

PhysicsSingle correct+4 / -1Question 11 of 90

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be :

Block M on a frictionless floor joined by a spring of constant k to a wall; a small mass m sits on top of M.

A\sqrt{\dfrac{M+m}M}

A\sqrt{\dfrac M{M+m}}

A\sqrt{\dfrac{M-m}M}

A\sqrt{\dfrac M{M-m}}

Answer and worked solution

Official answer: B

At the mean position momentum is conserved: $v'=\frac{Mv}{M+m}$ , and $\omega'=\sqrt{\frac k{M+m}}$ .

Amplitude $A'=\frac{v'}{\omega'}=\frac{M\,A\omega}{M+m}\sqrt{\frac{M+m}k}=A\sqrt{\frac M{M+m}}$ .

(B)

Topics:Spring-mass systems Conservation of linear momentum

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JEE Main 2021 · 24 Feb · Shift 1, Physics question 11