Two springs, of force constants $k_{1}$ and $k_{2}$, are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$. If both $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$ are made four times their original values, the frequency of oscillation becomes

JEE Main · 2007 · 29 Apr 2007 · Offline · Elasticity & SHM

Two springs, of force constants

k_{1}

and

k_{2}

, are connected to a mass

m

as shown. The frequency of oscillation of the mass is

f

. If both

k_{1}

and

k_{2}

are made four times their original values, the frequency of oscillation becomes

(A) $f /2$
(B) $f /4$
(C) $4 f$
(D) $2 f$ ✓ Correct

Answer: (D)

Solution

When two springs act together in the parallel arrangement shown, their restoring forces add for a common displacement, so the pair behaves like a single spring whose constant is the sum

k_{1} + k_{2}

. The frequency of the resulting simple harmonic motion is

f = \frac{1}{2 π} \frac{k _{1} + k _{2}}{m}

. If each constant is raised to four times its value the effective constant becomes

4 (k_{1} + k_{2})

, and since the frequency varies as the square root of the stiffness, the new frequency is

\frac{1}{2 π} \frac{4 ( k _{1} + k _{2} )}{m} = 2 f

. Quadrupling every stiffness therefore doubles the frequency of oscillation.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

Practise more: all JEE Main PYQs on Elasticity & SHM →

JEE Main · 2007 · 29 Apr 2007 · Offline · Elasticity & SHM

Solution

More JEE Main questions on Elasticity & SHM