Does the time period of a spring-mass system depend on g?

No. The period is T = 2π√(m/k) for both horizontal and vertical spring-mass systems. For a vertical spring, gravity only shifts the equilibrium downward by x₀ = mg/k; when the mass oscillates about that new equilibrium, gravity is exactly cancelled by the spring force at equilibrium and the net restoring force is again −ky. NIOS states it directly: acceleration due to gravity does not influence the vertical oscillations of a spring-mass system. The same spring on the Moon would oscillate with the same period (for the same m and k).

Does the period change with amplitude?

No. Because the restoring force F = −kx is exactly linear in displacement for an ideal spring, the motion is true simple harmonic motion and T = 2π√(m/k) is independent of amplitude. NCERT lists this as a Point to Ponder: the period of SHM does not depend on amplitude, energy or phase constant. Pulling the block 2 cm or 8 cm gives the same period; only the maximum speed and energy change.

Why does cutting a spring make it stiffer?

Force constant is inversely proportional to natural length, so k·L = constant for a given spring material. Cutting a spring of constant k into n equal pieces gives each piece a constant nk — each shorter piece stretches less for the same force, so it is stiffer. Cutting a spring into lengths in the ratio 1:2:3 gives constants 6k, 3k and 2k respectively, since each k_i is inversely proportional to its length fraction.

When are springs in series and when in parallel?

Springs are in series when the same force passes through both (joined end to end, so their extensions add): 1/k_eff = 1/k₁ + 1/k₂, and the combination is softer than either spring. Springs are in parallel when both share the same displacement (side by side or on opposite sides of the block): k_eff = k₁ + k₂, and the combination is stiffer. The period uses k_eff: T = 2π√(m/k_eff).

In NCERT Example 13.6, why is the period 2π√(m/2k) and not 2π√(m/k)?

Two identical springs of constant k are attached on opposite sides of the block to two fixed walls. When the block moves a distance x, one spring stretches by x and the other compresses by x, but both push the block back toward the mean position — so the forces add: F = −kx − kx = −2kx. This is a parallel combination with k_eff = 2k, giving T = 2π√(m/2k).

Does the mass of the spring affect the period?

In the NEET-level treatment the spring is taken to be massless, so the period is exactly T = 2π√(m/k). For a real spring of mass m_s, a more accurate result replaces m by (m + m_s/3) — the effective mass adds one-third of the spring's mass. NEET problems assume a massless spring unless explicitly stated; treat any 'spring of mass m_s' phrasing as a flag for the m_s/3 correction.

Spring-Mass System — NEET Physics

The restoring force and the period

A spring of force constant $k$ attached to a block of mass $m$ is the cleanest oscillator in mechanics. Displace the block by $x$ from its natural (relaxed) position and the spring pulls it back with a force $F=-kx$ — Hooke's law, the negative sign marking that the force opposes the displacement. This is precisely the SHM force law that the chapter on the force law for SHM derives from $a=-\omega^2 x$, with the identification $k=m\omega^2$.

Because the spring force is linear in $x$, the block executes true simple harmonic motion. Rearranging $k=m\omega^2$ gives the angular frequency $\omega=\sqrt{k/m}$, and since $T=2\pi/\omega$, the period is

$$ T = 2\pi\sqrt{\dfrac{m}{k}}, \qquad \omega = \sqrt{\dfrac{k}{m}}, \qquad \nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}. $$

NCERT lists this verbatim in its summary: a particle of mass $m$ oscillating under a Hooke's-law restoring force $F=-kx$ is a linear oscillator with $\omega=\sqrt{k/m}$ and $T=2\pi\sqrt{m/k}$. Two consequences are worth fixing at once, because NEET tests them year after year: a stiffer spring (larger $k$) oscillates faster, and a heavier block (larger $m$) oscillates slower. Neither $g$ nor the amplitude appears anywhere in the formula.

Horizontal spring-mass oscillator

Place the block on a frictionless horizontal surface, with the spring's far end clamped to a rigid wall. At equilibrium the spring is relaxed and the block sits at $x=0$. Pull it a small distance $x$ to the right: the spring stretches and exerts $kx$ to the left. Release it, and the restoring force drives the block back through the mean position. By inertia it overshoots, compresses the spring, is pushed back again, and the cycle repeats — exactly the to-and-fro motion NIOS §13.3.1 describes.

Figure 1 — Horizontal oscillator

The block is displaced a distance $x$ to the right of the equilibrium $x=0$. The stretched spring exerts a restoring force $F=-kx$ directed back toward the mean position. The motion is SHM with $T=2\pi\sqrt{m/k}$.

The mean position here coincides with the spring's relaxed length: with no gravity component along the motion, the only horizontal force is the spring force, which is zero when $x=0$. Apply Newton's second law along the motion, $m\,\ddot x = -kx$, i.e. $\ddot x = -(k/m)\,x$. Comparing with the SHM acceleration equation $\ddot x = -\omega^2 x$ gives $\omega^2=k/m$ and hence $T=2\pi\sqrt{m/k}$.

Why T is independent of g and amplitude

Two of the most reliable NEET distractors hide inside this single formula. Notice what is not in $T=2\pi\sqrt{m/k}$: there is no $g$ and no amplitude $A$. Both omissions are physical, not accidental.

The amplitude drops out because the spring force is exactly proportional to displacement. NCERT lists this as a Point to Ponder: "The period of SHM does not depend on amplitude or energy or the phase constant." Whether you pull the block 2 cm or 8 cm, the restoring force scales up in exact proportion, the acceleration scales with it, and the time to complete one cycle stays fixed. Larger amplitude only raises the maximum speed and the total energy — never the period. This contrasts sharply with planetary orbits, where the period does depend on the orbit size (Kepler's third law).

Vertical spring-mass oscillator

Hang the same spring vertically from a rigid support and attach the block to its lower end. Gravity now stretches the spring even before the block oscillates. The block settles at a new equilibrium where the upward spring force balances the weight: $k\,x_0 = mg$, giving an equilibrium extension $x_0 = mg/k$. NIOS §13.3.2 writes this directly as the relation $k=mg/l$ for the static extension $l$.

Figure 2 — Vertical oscillator

Gravity stretches the spring by $x_0=mg/k$ to a new equilibrium. About that point, a small extra displacement $y$ gives a net restoring force $-ky$: gravity is fully cancelled at equilibrium, so the period is again $T=2\pi\sqrt{m/k}$.

Now displace the block a further small distance $y$ below this new equilibrium. The spring is stretched by $x_0+y$, so it pulls up with force $k(x_0+y)$, while gravity pulls down with $mg$. The net force is

$$ F = mg - k(x_0+y) = mg - kx_0 - ky = -ky, $$

because $kx_0=mg$ cancels the weight exactly. The net restoring force about the new equilibrium is $-ky$ — the same elastic form as the horizontal case. Hence $\omega=\sqrt{k/m}$ and $T=2\pi\sqrt{m/k}$ once more. NIOS states the conclusion plainly: acceleration due to gravity does not influence the vertical oscillations of a spring-mass system. Gravity sets where the block oscillates, not how fast.

Horizontal spring

Mean position = spring's natural length.
Only horizontal force is the spring force $-kx$.
Gravity acts perpendicular to the motion (balanced by the normal reaction) — it never enters the equation.
$T = 2\pi\sqrt{m/k}$.

Vertical spring

Mean position = stretched equilibrium, $x_0=mg/k$ below the natural length.
Two forces: weight $mg$ down and spring $k(x_0+y)$ up.
At equilibrium they cancel; net restoring force about it is $-ky$, so gravity drops out of $T$.
$T = 2\pi\sqrt{m/k}$ — identical to the horizontal case.

Related drill

The spring stores energy $U=\tfrac12 kx^2$ that swaps with kinetic energy each cycle — work through energy in SHM for the full $E=\tfrac12 kA^2$ picture.

Combinations of springs

NEET frequently replaces the single spring with two springs and asks for the new period. The whole task reduces to finding one effective force constant $k_{\text{eff}}$ and substituting it into $T=2\pi\sqrt{m/k_{\text{eff}}}$. Two arrangements cover almost every problem.

Figure 3 — Series vs parallel

Series: the springs join end to end, the same tension passes through both, their extensions add, and the combination is softer ($1/k_{\text{eff}}=1/k_1+1/k_2$). Parallel: both springs share the block's displacement, their forces add, and the combination is stiffer ($k_{\text{eff}}=k_1+k_2$).

Springs in series

When two springs are joined end to end, the same force $F$ is transmitted through both (the tension is uniform along the chain). Each stretches independently: $x_1=F/k_1$ and $x_2=F/k_2$. The total extension is the sum, $x=x_1+x_2=F\left(1/k_1+1/k_2\right)$. Writing $x=F/k_{\text{eff}}$ gives

$$ \frac{1}{k_{\text{eff}}}=\frac{1}{k_1}+\frac{1}{k_2}, \qquad k_{\text{eff}}=\frac{k_1 k_2}{k_1+k_2}. $$

The series constant is always smaller than either individual constant — the chain is softer, so the period $T=2\pi\sqrt{m/k_{\text{eff}}}$ is longer.

Springs in parallel

When two springs both connect the block to fixed supports so that they share the same displacement $x$, each exerts its own force and the forces add: $F=k_1 x + k_2 x = (k_1+k_2)x$. Hence

$$ k_{\text{eff}}=k_1+k_2. $$

The parallel constant is always larger than either individual constant — the combination is stiffer, so the period is shorter. This is exactly the situation in NCERT Example 13.6, where two identical springs on opposite sides of the block give $k_{\text{eff}}=2k$.

Cutting a spring changes k

The force constant of a spring is inversely proportional to its natural length: for a uniform spring, the product $k\,L$ is a constant of the material and coil geometry. Physically, a longer spring has more coils sharing the load, so each coil stretches less and the spring as a whole yields more for a given force — a smaller $k$. Cut the spring shorter and $k$ goes up.

Cut a spring of constant $k$ into $n$ equal pieces and each piece has length $L/n$, so each has constant $nk$. Cut it into unequal lengths in the ratio $\ell_1:\ell_2:\ell_3$ and each piece's constant is inversely proportional to its length fraction. For a cut in the ratio $1:2:3$ (total $6$ parts), the three pieces have constants $6k,\,3k,\,2k$ respectively — the shortest piece is the stiffest. This is the exact engine of NEET 2017 Q.180, solved below.

NCERT spring examples

NCERT Example 13.6

Two identical springs of spring constant $k$ are attached to a block of mass $m$ and to fixed supports on either side. Show that displacing the mass executes SHM, and find the period.

Set-up. Displace the block a small distance $x$ to the right. The spring on the left is elongated by $x$ and exerts $F_1=-kx$ (pulling the block back toward the mean position). The spring on the right is compressed by $x$ and exerts $F_2=-kx$ (pushing the block back). Both forces point the same way — toward the mean position.

Net force. $F=F_1+F_2=-kx-kx=-2kx$. The force is proportional to displacement and directed toward the mean position, so the motion is simple harmonic. This is a parallel combination with $k_{\text{eff}}=2k$.

Period. $T=2\pi\sqrt{\dfrac{m}{2k}}$, as NCERT states.

NCERT Example 13.9 (Exercise 13.9)

A spring of force constant $1200~\text{N m}^{-1}$ is mounted on a horizontal table. A mass of $3~\text{kg}$ is attached, pulled $2.0~\text{cm}$ aside and released. Find the frequency, the maximum acceleration and the maximum speed.

Angular frequency. $\omega=\sqrt{k/m}=\sqrt{1200/3}=\sqrt{400}=20~\text{rad s}^{-1}$. Frequency $\nu=\omega/2\pi=20/6.28\approx 3.2~\text{Hz}$.

Maximum acceleration. $a_{\max}=\omega^2 A=(20)^2\times 0.02=400\times0.02=8~\text{m s}^{-2}$.

Maximum speed. $v_{\max}=\omega A=20\times0.02=0.4~\text{m s}^{-1}$. The amplitude $A=2~\text{cm}$ enters the maxima but not the period — exactly the amplitude-independence above.

NCERT Example 13.13 (Exercise 13.13)

(a) A spring of constant $k$ is clamped at one end with mass $m$ on the free end; a force $F$ stretches it. (b) The same spring has both ends free, with a mass $m$ at each end, each end stretched by the same force $F$. Compare the maximum extension and the period of oscillation in the two cases.

Maximum extension. In both cases the spring experiences the same tension $F$ throughout (a spring is in tension uniformly), so the extension is $x=F/k$ in both arrangements. The two free ends in (b) do not double the extension; the wall in (a) simply plays the role one of the masses plays in (b).

Period. In (a) the single mass $m$ oscillates against a spring of constant $k$: $T=2\pi\sqrt{m/k}$. In (b) each mass oscillates about the centre of mass; by symmetry the midpoint is stationary, so each half-spring (constant $2k$) drives a mass $m$. The period is therefore $T=2\pi\sqrt{m/2k}$ — but the standard NCERT answer treats the two-mass system with the reduced mass $\mu=m/2$ against the full spring $k$, again giving $T=2\pi\sqrt{m/2k}$. The free-free oscillator is faster than the clamped one.

Quick recap

Spring-mass system in one breath

Restoring force $F=-kx$ (Hooke) → $\omega=\sqrt{k/m}$ and $T=2\pi\sqrt{m/k}$.
Period is independent of $g$ and of amplitude — unlike the pendulum, the spring oscillates the same on the Moon.
Vertical spring: gravity only shifts equilibrium by $x_0=mg/k$; net restoring force about it is $-ky$, so $T=2\pi\sqrt{m/k}$ unchanged.
Series: $1/k_{\text{eff}}=1/k_1+1/k_2$ (softer, longer $T$). Parallel: $k_{\text{eff}}=k_1+k_2$ (stiffer, shorter $T$).
NCERT Ex. 13.6: two springs on opposite sides → parallel, $k_{\text{eff}}=2k$, $T=2\pi\sqrt{m/2k}$.
Cutting a spring raises $k$: each piece's constant is inversely proportional to its length; $n$ equal parts → $nk$ each.

Spring-Mass System