If spring mass system has frequency v₁ and spring is cut to half, what is v₂/v₁?

√2. Cutting spring to half doubles k (k∝1/L). Frequency ∝ √k, so v₂/v₁=√(2k/k)=√2.

Why does halving spring length double spring constant?

k=YA/L (Young's modulus × area / length). Halving L gives k'=2k.

Spring mass system frequency v₁, spring cut to half — new frequency v₂ — find v₂/v₁

JEERANKERS

PhysicsSHM

QThe frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ is:

A) $1$ B) $2$ C) $\sqrt{2}$ D) $\sqrt{3}$Simple Harmonic Motion

MCQ

✅ Correct Answer

C) √2

✓

Solution

Spring constant when cut to half

Spring constant $k$ is inversely proportional to length: $k\propto1/L$.

Cutting spring to half length: new spring constant $k'=2k$.

Frequency ratio

$v=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$\dfrac{v_2}{v_1}=\sqrt{\dfrac{k'}{k}}=\sqrt{\dfrac{2k}{k}}=\sqrt{2}$

📘 Spring Constant and Length

Spring constant $k=YA/L$ (Young's modulus × area / length). Halving the length doubles $k$. Frequency $v=(1/2\pi)\sqrt{k/m}\propto\sqrt{k}$. So $v_2/v_1=\sqrt{2k/k}=\sqrt{2}$.

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Important Concepts

Spring Constant Inversely Proportional to LengthFor a spring of natural length L: k=YA/L. When cut to L/2: k'=YA/(L/2)=2k. The shorter spring is stiffer.

Frequency of Spring-Mass Systemf=(1/2π)√(k/m). Frequency depends on √k. Doubling k increases frequency by √2.

Physical IntuitionA shorter spring is stiffer (less extension per unit force). Stiffer spring → faster oscillation. Half length → twice k → frequency increases by √2.

What if spring cut to 1/n?k→nk. Frequency ratio = √n. For half: √2 ≈ 1.414.

FAQs

Why does cutting spring to half double k?

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k=YA/L. Halving L means k'=YA/(L/2)=2YA/L=2k. Think of it as two springs in series: each half has k'=2k, and in series they give k_total=k'k'/(k'+k')=k'/2=k — the original spring. So each half must have k'=2k.

Does the mass of the spring matter?

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For ideal massless spring: no. In reality, spring mass affects frequency, but JEE problems use ideal springs.

What is the new frequency in Hz?

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v₂=√2·v₁≈1.414·v₁. If original was 1 Hz, new is 1.414 Hz.

What if instead the spring is cut to 1/3?

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k→3k. v₂=√3·v₁.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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