MHT-CET Physics · Oscillations

Spring-Mass Systems, Spring Combinations and Other Oscillators

A mass m on a spring of constant k oscillates with T = 2π√(m/k), whether the spring is horizontal or vertical; springs combine like capacitors (parallel constants add, series reciprocals add), cutting a spring raises its constant, and any system with a restoring force proportional to displacement — a floating block, a liquid in a U-tube, a magnetic needle — has a period of the same form.

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Why this matters

25 PYQs, 7 HARD. Twelve are one spring and a changing mass — the period with 4m, the mass that stretches T to 5T/4, the static stretch from the period, a smaller mass dropped on at the mean position; eight are springs combined or cut — two in parallel under a disc, springs in series and parallel on both sides of a block, a spring cut in two, and a stretched spring whirled in a circle; five are other oscillators — a ball in a bowl, a floating block, a liquid column, a magnetic needle. Three cards.

Concept 1 of 3: The Spring-Mass Period

T = 2π√(m/k): the period grows as the square root of the mass. So 4m doubles T, and if adding m₀ stretches T to 5T/4, then (m + m₀)/m = 25/16 and m₀/m = 9/16. A hanging mass stretches the spring by x = mg/k at rest, so T = 2π√(x/g) — the period tells you the stretch without knowing m or k. A mass dropped gently onto an oscillating one at the mean position conserves momentum, and the new amplitude follows from Mω₁A₁ = (M + m)ω₂A₂.

Definition

$T = 2\pi\sqrt{\dfrac{m}{k}}$ , $T \propto \sqrt{m}$ : 4m ⇒ 2T.
Adding mass: $\dfrac{m + m_0}{m} = \left(\dfrac{T'}{T}\right)^2$ (T → 5T/4 ⇒ $\tfrac{9}{16}$ ; T → 4T/3 ⇒ $\tfrac{7}{9}$ ; 3 s → 5 s with 1 kg ⇒ m = 9/16 kg).
Static stretch $x = \dfrac{mg}{k} = \dfrac{gT^2}{4\pi^2}$ (T = 6 s, g = π² ⇒ 9 m). A mass $M_1$ that adds stretch x gives $k = \dfrac{M_1g}{x}$ .
Mass dropped on at the mean position: $\dfrac{A_1}{A_2} = \sqrt{\dfrac{M + m}{M}}$ .
Speed at displacement x: $\sqrt{\tfrac{k}{m}}\sqrt{A^2 - x^2}$ .

Spring-mass

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

Worked example

A 0.5 kg mass on a spring has period 1 s. How much extra mass makes the period 1.5 s?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q21Moderate

Example 1 · Oscillations · Spring-Mass and Other SHM Systems

The upper end of the spring is fixed and a mass 'm' is attached to its lower end. When mass is slightly pulled down and released, it oscillates with time period 3 second. If mass 'm' is increased by 1 kg, the time period becomes 5 second. The value of 'm' is (mass of spring is negligible)

Scaling the period with the mass itself

T grows as √m. Doubling the mass makes the period √2 times, not twice — the ratios in these questions are always squared.

Putting g into a spring's period

Hanging the spring vertically only moves the equilibrium point down by mg/k; the period is still 2π√(m/k). g enters only when the question gives you the static stretch instead of m and k.

Drill 11 more on the spring-mass period

Concept 2 of 3: Springs in Series, in Parallel, and Cut

Two springs side by side both stretch the same amount, so their constants add: parallel k = k₁ + k₂. Two end to end share the force and add their stretches: series 1/k = 1/k₁ + 1/k₂. A block between two springs fixed to opposite walls is pushed by one and pulled by the other — also k₁ + k₂. A spring's constant is inversely proportional to its length, so cutting it into a piece of length l₁ gives k l/l₁, and half a spring is twice as stiff.

Definition

Parallel (and a block between two walls): $k = k_1 + k_2$ . Series: $\dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2}$ .
Cut spring: $kl$ constant; half ⇒ 2k, period ÷ √2. Cut into $l_1 = nl_2$ : $k_1 = \dfrac{(n+1)k}{n}$ .
Identical springs K: single $T_a$ , series $\sqrt{2}T_a$ , parallel $\tfrac{T_a}{\sqrt{2}}$ — so $T_b = 2T_c$ .
Disc of 12 kg on two springs, T = 2 s, π² = 10 ⇒ each spring 60 N/m.
A spring whirled in a circle: $kx = m\omega^2(l + x)$ ⇒ $\dfrac{x}{l} = \dfrac{m\omega^2}{k - m\omega^2}$ .

Combining springs

k_{\parallel} = k_1 + k_2, \qquad \frac{1}{k_{\text{series}}} = \frac{1}{k_1} + \frac{1}{k_2}

Worked example

A block rests between a 3K spring on the left and two K springs in series on the right, all fixed to walls. Frequency, mass M?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q23Hard

Example 2 · Oscillations · Spring-Mass and Other SHM Systems

All the springs in fig. (a), (b) and (c) are identical, each having force constant K. Mass attached to each system is 'm'. If

T_a, T_b

and

T_c

are the time periods of oscillations of the three systems respectively, then

Treating springs on opposite sides as series

A block between two walls stretches one spring and compresses the other by the SAME x, so both forces act on it: the constants add, as in parallel.

Drill 7 more on springs in series, in parallel, and cut

Concept 3 of 3: Other Oscillators: Bowls, Floating Blocks, Liquid Columns, Magnets

Find the restoring force per unit displacement and the same T = 2π√(inertia/restoring constant) appears. A small ball in a smooth bowl of radius R behaves like a pendulum of length R (or R − r, measuring to the ball's centre). A floating block pushed down by x gets an extra upthrust Aρgx, so ω² = Aρg/m, and for a block floating with a height da immersed, T = 2π√(da/g). A liquid of mass M in a U-tube displaced by y has a level difference 2y, a restoring force 2Aydg, and T = 2π√(M/(2Adg)). A magnetic needle has T = 2π√(I/mB).

Definition

Bowl / watch glass of radius R: $T = 2\pi\sqrt{\dfrac{R}{g}}$ (1.6 m ⇒ 0.8π s); a rolling ball of radius r: $\propto \sqrt{\dfrac{R - r}{g}}$ .
Floating block: $n = \dfrac{1}{2\pi}\sqrt{\dfrac{A\rho g}{m}}$ ; wood of relative density d with side a vertical: $T = 2\pi\sqrt{\dfrac{ad}{g}}$ .
Liquid column (U-tube), mass M: $T = 2\pi\sqrt{\dfrac{M}{2Adg}}$ .
Magnetic needle: $T = 2\pi\sqrt{\dfrac{I}{mB}}$ .

Liquid column and floating block

T_{\text{column}} = 2\pi\sqrt{\frac{M}{2Adg}}, \qquad T_{\text{block}} = 2\pi\sqrt{\frac{ad}{g}}

Worked example

A cube of side 20 cm and relative density 0.6 floats in water. Period of small vertical oscillations (g = π²)?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q10Hard

Example 3 · Oscillations · Spring-Mass and Other SHM Systems

A piece of wood has length, breadth and height, 'a', 'b' and 'c' respectively. Its relative density is 'd'. It is floating in water such that the side 'a' is vertical. It is pushed down a little and released. The time period of S.H.M. executed by it is (

g =

acceleration due to gravity)

Using Adg for a liquid column

Displacing the liquid by y lowers one side by y and raises the other by y, a level difference of 2y. The restoring force is 2Aydg, which puts the 2 inside the square root.

Drill 4 more on other oscillators: bowls, floating blocks, liquid columns, magnets

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The Spring-Mass Period
Spring-mass
$T = 2\pi\sqrt{\frac{m}{k}}$
Springs in Series, in Parallel, and Cut
Combining springs
$k_{\parallel} = k_1 + k_2, \qquad \frac{1}{k_{\text{series}}} = \frac{1}{k_1} + \frac{1}{k_2}$
Other Oscillators: Bowls, Floating Blocks, Liquid Columns, Magnets
Liquid column and floating block
$T_{\text{column}} = 2\pi\sqrt{\frac{M}{2Adg}}, \qquad T_{\text{block}} = 2\pi\sqrt{\frac{ad}{g}}$

Watch out for (4)

Scaling the period with the mass itself
→ The Spring-Mass Period
Putting g into a spring's period
→ The Spring-Mass Period
Treating springs on opposite sides as series
→ Springs in Series, in Parallel, and Cut
Using Adg for a liquid column
→ Other Oscillators: Bowls, Floating Blocks, Liquid Columns, Magnets

Test yourself on Oscillations

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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