(a) Describe the horizontal oscillations of a spring. **OR** (b) Describe rolling on inclined plane and arrive

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Q. (a) Describe the horizontal oscillations of a spring. OR

(b) Describe rolling on inclined plane and arrive at the expression for the acceleration.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2025Subjective· 5mImportance★★★★★

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Concept understanding — Spring-Mass System Period

The Spring-Mass System: Why a Weight on a Spring Oscillates

Imagine hanging a weight from a spring and giving it a gentle tug downward. It bounces back up, overshoots, comes down again, and keeps going. That rhythmic up-and-down motion is simple harmonic motion, and the time it takes to complete one full bounce — down, up, and back to the start — is called the period $T$ .

The key question: what determines how fast or slow this bouncing happens? Intuition says two things matter: how heavy the weight is, and how stiff the spring is.

The Intuition

A heavier mass is harder to accelerate. It lumbers along, so the oscillation is slower — the period gets longer. A stiffer spring (one with a larger spring constant $k$ ) pulls back harder for the same stretch. That stronger restoring force whips the mass back faster, so the period gets shorter.

So the period $T$ should increase with mass $m$ and decrease with stiffness $k$ . The exact relationship turns out to be:

T = 2\pi \sqrt{\frac{m}{k}}

The $2\pi$ factor comes from the geometry of circular motion (which underlies all simple harmonic motion), and the square root tells us that doubling the mass only multiplies the period by $\sqrt{2} \approx 1.4$ , not by 2.

Where Does This Formula Come From?

For a mass on an ideal spring, the restoring force is Hooke's law: $F = -kx$ , where $x$ is the displacement from equilibrium. Newton's second law $F = ma$ gives:

m \frac{d^2x}{dt^2} = -kx

This is a differential equation whose solution is a sine or cosine wave. The angular frequency $\omega$ (how fast the oscillation goes in radians per second) comes out as:

\omega = \sqrt{\frac{k}{m}}

Since period $T$ is the time for one complete cycle, and one cycle corresponds to $2\pi$ radians, we have $T = 2\pi / \omega$ . Substituting $\omega$ gives the formula above.

Tip

If you ever forget which variable goes in the numerator, remember: mass in the numerator makes the period longer (heavier = slower), and stiffness in the denominator makes the period shorter (stiffer = faster). The square root just moderates the effect.

What the Formula Tells You

Mass and period: Double the mass → period increases by $\sqrt{2}$ (about 1.4 times). Quadruple the mass → period doubles.
Stiffness and period: Double the spring constant → period decreases by $\sqrt{2}$ (about 0.7 times). Quadruple the stiffness → period halves.
Independence from amplitude: The period does not depend on how far you pull the mass initially. A small bounce and a big bounce take exactly the same time. This is the hallmark of simple harmonic motion for an ideal spring. …

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A horizontal spring-mass system on a frictionless surface obeys Hooke's Law (F=-kx), leading to SHM with T = 2pisqrt(m/k). (This question offered an internal choice; part (a), horizontal spring oscillations, is answered here.)

Setup: Consider a block of mass m attached to one end of a spring of force constant (stiffness) k; the other end of the spring is fixed to a wall, and the block rests on a frictionless horizontal surface. Let x be the displacement of the block from its equilibrium (natural, unstretched) position.

Restoring force: By Hooke's Law, when the spring is stretched or compressed by x from its natural length, it exerts a restoring force on the block that is proportional to x and directed opposite to the displacement (always pushing/pulling the block back toward equilibrium):

F = -k x

(the negative sign shows the force always opposes the displacement).

Equation of motion: Applying Newton's Second Law, F = m (d^2x/dt^2):

m (d^2x/dt^2) = -k x

d^2x/dt^2 = -(k/m) x

This is exactly the standard differential equation of Simple Harmonic Motion, d^2x/dt^2 = -omega^2 x, with:

omega^2 = k/m, so omega = sqrt(k/m)

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The angular frequency of a body of mass m suspended with a spring of spring constant k is -$
(a) sqrt(m/k)
(b) sqrt(k/m)
(c) (1/2) m x k
(d) k/m
›Reveal solutionSolution

For a spring-mass system, ω = sqrt(k/m), derived by comparing Newton's second law with the SHM differential equation.

For a mass m displaced by x from equilibrium on a spring of spring constant k, the restoring force is F = -kx (Hooke's law). By Newton's second law:

m (d²x/dt²) = -kx

→ (d²x/dt²) = -(k/m) x

The standard equation for simple harmonic motion is (d²x/dt²) = -ω² x. Comparing the two:

ω² = k/m

ω = sqrt(k/m)

…

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CBSE 2025Set ANNUAL1 mark
$Q. What will be the change in time period of a spring pendulum when taken to the moon?$
›Reveal solutionSolution

A spring (mass-spring system) pendulum's restoring force comes from the spring constant, not from gravity, so its period is independent of the local value of $g$ — unlike a simple pendulum.

For a mass $m$ attached to a spring of force constant $k$ , oscillating horizontally or vertically, the restoring force is $F=-kx$ , and Newton's second law gives:

$T = 2\pi\sqrt{\dfrac{m}{k}}$

This expression contains only the mass $m$ and the spring constant $k$ ; it does not contain the acceleration due to gravity $g$ at all (gravity only shifts the equilibrium position for a vertical spring, it does not change the restoring-force constant).

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. A mass-spring system is oscillating in a car. If the car moves on a horizontal road with accelerated speed, then its frequency (A) will decrease (B) will increase (C) will remain same (D) will be zero$
›Reveal solutionSolution

A mass-spring system's oscillation frequency is unaffected by the accelerating frame it's placed in.

The frequency of a spring-mass oscillator is $f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ , which depends only on the spring constant $k$ and the oscillating mass $m$ . If the car accelerates, a pseudo-force acts on the system in its (non-inertial) frame, but this only shifts the equilibrium (mean) posit …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Two bodies A and B whose masses are in the ratio 1 : 2 are suspended from two separate massless springs of force constants kA and kB respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio 1 : 2, the ratio of the amplitude A to that of B is ____.$
(a) sqrt(2 kB / kA)
(b) sqrt(kB / 2 kA)
(c) sqrt(8 kB / kA)
(d) sqrt(kB / 8 kA)
›Reveal solutionSolution

Using v_max = A√(k/m) for each body and the given mass ratio (1:2) and velocity ratio (1:2), the amplitude ratio works out to sqrt(kB/8kA).

For a mass m attached to a spring of constant k, undergoing SHM with amplitude A, the maximum velocity is:

v_max = A ω = A √(k/m)

Let mass of A be m, so mass of B = 2m (given mA : mB = 1 : 2).

Given v_maxA : v_maxB = 1 : 2, i.e. v_maxA / v_maxB = 1/2.

Write:

v_maxA = AA √(kA/m)

v_maxB = AB √(kB/2m)

So: …

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CBSE 2022Set ANNUAL1 mark
$Q. Write the unit of spring constant.$
›Reveal solutionSolution

The spring constant k, defined by F = -kx, has SI unit N/m (equivalently kg/s²).

Hooke's law for a spring states that the restoring force is F = -kx, where x is the displacement from the natural length and k is the spring (force) constant. Rearranging, k = -F/x, so its unit is [Force]/[Length] = N/m. This same constant …

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Q.(a) Describe the horizontal oscillations of a spring. OR

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