JEE Main · 2004 · 24 Apr 2004 · Offline · Elasticity & SHM

A particle at the end of a spring executes simple harmonic motion with a period

t_{1}

, while the corresponding period for another spring is

t_{2}

. If the period of oscillation with the two springs in series is

t

, then

Answer: (B)

Solution

For two springs joined in series the compliances add,

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

. Since a period satisfies

T = 2 π \frac{m}{k}

, we have

k = \frac{4 π ^{2} m}{T ^{2}}

, so

\frac{1}{k}

is proportional to

T^{2}

. Putting the three constants into the series relation gives

T^{2} = t_{1}^{2} + t_{2}^{2}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option