Oscillations and SHM: Step-by-Step Worked Examples (2)

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Tags Oscillations

Question

A spring of force constant k=200k = 200 N/m is cut into two equal halves. One half is attached to a 2 kg block on a frictionless surface and the block is pulled 5 cm from equilibrium and released. Find: (a) the new spring constant of the half-spring, (b) the angular frequency of oscillation, (c) the maximum speed of the block.

Solution — Step by Step

When a spring is cut into two equal pieces, each half has double the original stiffness. Reason: the same applied force now stretches half the original length, so k′=2k=400k' = 2k = 400 N/m. This is a JEE Main favourite — half spring, double kk.

For a mass-spring system:

ω=k′m=4002=200≈14.14 rad/s\omega = \sqrt{\frac{k'}{m}} = \sqrt{\frac{400}{2}} = \sqrt{200} \approx 14.14 \text{ rad/s}

Maximum speed in SHM is vmax⁡=Aωv_{\max} = A \omega, where A=0.05A = 0.05 m (5 cm).

vmax⁡=0.05⋅14.14≈0.707 m/sv_{\max} = 0.05 \cdot 14.14 \approx 0.707 \text{ m/s}

(a) k′=400k' = 400 N/m, (b) ω≈14.14\omega \approx 14.14 rad/s, (c) vmax⁡≈0.707v_{\max} \approx 0.707 m/s.

Why This Works

A spring’s kk depends on how much it stretches per unit force. Halving the length halves the stretch for the same force, so stiffness doubles. Mathematically: k∝1/Lk \propto 1/L for identical material and cross-section.

Once kk is known, SHM formulas ω=k/m\omega = \sqrt{k/m}, vmax⁡=Aωv_{\max} = A\omega, amax⁡=Aω2a_{\max} = A\omega^2, T=2πm/kT = 2\pi\sqrt{m/k} flow directly.

If a spring is cut in ratio 1:21:2, the longer piece has kk scaled by 3/23/2 and the shorter by 33. General rule: kpiece=koriginal⋅(Loriginal/Lpiece)k_{\text{piece}} = k_{\text{original}} \cdot (L_{\text{original}}/L_{\text{piece}}).

Alternative Method

Use energy conservation for vmax⁡v_{\max}: at maximum speed, all PE has converted to KE. So 12k′A2=12mvmax⁡2\frac{1}{2}k'A^2 = \frac{1}{2}mv_{\max}^2, giving vmax⁡=Ak′/m=0.05⋅200=0.707v_{\max} = A\sqrt{k'/m} = 0.05 \cdot \sqrt{200} = 0.707 m/s. Same answer without explicit ω\omega.

Common Mistake

Students often use k=200k = 200 (the original) for the half-spring period, getting ω=10\omega = 10 rad/s and vmax⁡=0.5v_{\max} = 0.5 m/s. Always update kk before plugging into ω=k/m\omega = \sqrt{k/m}.

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