JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Elasticity & SHM

The mass

M

shown in the figure oscillates in simple harmonic motion with amplitude

A

. The amplitude of the point P is

(A) $\frac{k _{1} A}{k _{2}}$
(B) $\frac{k _{2} A}{k _{1}}$
(C) $\frac{k _{1} A}{k _{1} + k _{2}}$
(D) $\frac{k _{2} A}{k _{1} + k _{2}}$ ✓ Correct

Answer: (D)

Solution

Let the extension in spring

k_{1}

x_{1}

and in spring

k_{2}

x_{2}

. Since the springs are in series, the total displacement is

x_{1} + x_{2} = A

. At equilibrium, the force in both springs is equal:

k_{1} x_{1} = k_{2} x_{2}

. Solving these two equations:

x_{1} = \frac{k _{2} A}{k _{1} + k _{2}}

. Point P is at the junction of the two springs, so its amplitude is

x_{1} = \frac{k _{2} A}{k _{1} + k _{2}}

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Single correct option

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