SHM — find time period of spring-mass system with two springs in parallel

Question

A block of mass $m$ is attached to two springs of spring constants $k_1$ and $k_2$ connected in parallel. Find the time period of oscillation.

(JEE Main 2021, similar pattern)

Solution — Step by Step

In a parallel combination, both springs stretch by the same displacement $x$ , and their restoring forces add up:

F = -k_1 x - k_2 x = -(k_1 + k_2)x

So the effective spring constant is:

k_{eff} = k_1 + k_2

ma = -(k_1 + k_2)x

a = -\frac{(k_1 + k_2)}{m}x = -\omega^2 x

This is SHM with $\omega^2 = \frac{k_1 + k_2}{m}$ .

T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k_{eff}}}

\boxed{T = 2\pi\sqrt{\frac{m}{k_1 + k_2}}}

Why This Works

“Parallel” means both springs share the same displacement but contribute independently to the force. Think of it as the block being pulled back by two springs simultaneously — both try to restore it to equilibrium, so the total restoring force is stronger.

A stronger effective spring means a higher frequency (faster oscillation) and shorter time period. This makes physical sense: two springs pulling together are stiffer than one alone.

For series springs, the result is different: $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$ , giving $k_{eff} = \frac{k_1 k_2}{k_1 + k_2}$ . Series springs are softer (smaller $k_{eff}$ ), so the time period is longer.

Alternative Method — Energy approach

Total PE when displaced by $x$ : $U = \frac{1}{2}k_1 x^2 + \frac{1}{2}k_2 x^2 = \frac{1}{2}(k_1 + k_2)x^2$

Total energy: $E = \frac{1}{2}mv^2 + \frac{1}{2}(k_1 + k_2)x^2 = \text{constant}$

Differentiating: $mv\frac{dv}{dt} + (k_1 + k_2)x\frac{dx}{dt} = 0$

Since $v = dx/dt$ : $ma + (k_1 + k_2)x = 0$ , confirming SHM with $\omega^2 = (k_1+k_2)/m$ .

For JEE, remember the spring combination rules (they’re opposite to capacitors and same as resistors): parallel springs add directly ( $k_{eff} = k_1 + k_2$ ), series springs add reciprocally ( $1/k_{eff} = 1/k_1 + 1/k_2$ ). Many MCQs test whether you know which rule applies.

Common Mistake

Students confuse parallel and series combinations. The key test: in parallel, both springs have the same displacement; in series, both springs carry the same force. If a block is sandwiched between two springs (one on each side), that’s parallel — both compress/extend by the same amount $x$ . If two springs are connected end-to-end with the block at one end, that’s series.

Question

Solution — Step by Step

Why This Works

Alternative Method — Energy approach

Common Mistake

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