JEE MainPhysics

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Oscillations and Waves: JEE Main Physics Question with Solution

Q: Why do the two springs act in parallel when they are on opposite sides of the mass?

Because displacing the mass by produces the same magnitude of displacement in both springs, and both exert restoring forces in the same direction (towards the equilibrium position). Thus, their effective stiffnesses add directly: .

A mass

\mathrm{m}

is attached to two springs as shown in figure. The spring constants of two springs are

\mathrm{K}_1

and

\mathrm{K}_2

. For the frictionless surface, the time period of oscillation of mass

\mathrm{m}

Mass m placed between two walls, attached to two springs of constants K2 and K1 on horizontal frictionless surface.

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\dfrac{1}{2\pi}\sqrt{\dfrac{\mathrm{K}_1 + \mathrm{K}_2}{\mathrm{m}}}

\dfrac{1}{2\pi}\sqrt{\dfrac{\mathrm{K}_1 - \mathrm{K}_2}{\mathrm{m}}}

2\pi \sqrt{\dfrac{\mathrm{m}}{\mathrm{K}_1 + \mathrm{K}_2}}

2\pi \sqrt{\dfrac{\mathrm{m}}{\mathrm{K}_1 - \mathrm{K}_2}}

Step-by-step solutionView

Correct answer

Displacing the mass by $x$ compresses one spring and elongates the other, giving a total restoring force of $F = -(K_1 + K_2)x$ , which yields an equivalent spring constant of $K_{\text{eq}} = K_1 + K_2$ and a time period of $T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .

Option analysis

Why each option works or fails

A ·

\dfrac{1}{2\pi}\sqrt{\dfrac{\mathrm{K}_1 + \mathrm{K}_2}{\mathrm{m}}}

Confusing the formula for natural linear frequency $f = \frac{1}{2\pi}\sqrt{\frac{K_{\text{eq}}}{m}}$ with the time period $T = \frac{1}{f}$ . Remember that time period has units of time (seconds), so it must be proportional to $\sqrt{\frac{m}{K_{\text{eq}}}}$ , giving $T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .

B ·

\dfrac{1}{2\pi}\sqrt{\dfrac{\mathrm{K}_1 - \mathrm{K}_2}{\mathrm{m}}}

Mistakenly subtracting the spring constants thinking the two opposing springs push against each other, while also inverting the formula to frequency instead of period. When the mass is displaced, both springs exert restoring forces in the same direction, so their spring constants add ( $K_{\text{eq}} = K_1 + K_2$ ), and the time period is $T = 2\pi\sqrt{m/K_{\text{eq}}}$ .

C ·

2\pi \sqrt{\dfrac{\mathrm{m}}{\mathrm{K}_1 + \mathrm{K}_2}}

None. This is the correct calculation. Both springs act in parallel because displacing the block by $x$ produces a combined restoring force $-(K_1 + K_2)x$ , giving $T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .

D ·

2\pi \sqrt{\dfrac{\mathrm{m}}{\mathrm{K}_1 - \mathrm{K}_2}}

Assuming that because the springs are on opposite sides of the mass, their restoring forces oppose each other and subtract. A displacement to the right compresses the right spring and stretches the left spring; both resulting forces push and pull the mass to the left, adding together to give $K_{\text{eq}} = K_1 + K_2$ .

Reviewed route

Solution

StepWorking

01given

Mass $m$ , spring constants $K_1$ and $K_2$ , frictionless surface.

02find

Time period of oscillation $T$ of mass $m$ .

03visualise

A mass $m$ is placed on a frictionless horizontal floor, attached to a fixed wall on the left via spring $K_1$ and a fixed wall on the right via spring $K_2$ . When displaced to the right by $x$ , the left spring is stretched by $x$ and pulls left with force $K_1 x$ ; the right spring is compressed by $x$ and pushes left with force $K_2 x$ .

04strategise

Find the net restoring force on the mass for a displacement $x$ : $F_{\text{net}} = -(K_1 + K_2)x$ . Relate this to $F = -K_{\text{eq}}x$ to find $K_{\text{eq}} = K_1 + K_2$ , then apply $T = 2\pi \sqrt{m/K_{\text{eq}}}$ .

05execute

Acceleration $a = \frac{F_{\text{net}}}{m} = -\left(\frac{K_1 + K_2}{m}\right)x = -\omega^2 x$ , yielding $\omega = \sqrt{\frac{K_1 + K_2}{m}}$ . Thus, $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .

06verify

Dimensionally $[T] = \text{s}$ . When $K_2 = 0$ , $T = 2\pi\sqrt{m/K_1}$ , which correctly matches a single spring-mass system.

07given

Mass $m$ connected between two rigid walls with springs $K_1$ and $K_2$ .

08find

Time period $T$ .

09strategise

When a mass is sandwiched between two springs attached to fixed supports, both springs experience identical displacement $x$ . This is a parallel combination: $K_{\text{eq}} = K_1 + K_2$ .

10execute

Substitute $K_{\text{eq}}$ into $T = 2\pi \sqrt{\frac{m}{K_{\text{eq}}}} = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .

✓verify

Directly matches Option (2).

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why do the two springs act in parallel when they are on opposite sides of the mass?

Because displacing the mass by $x$ produces the same magnitude of displacement $x$ in both springs, and both exert restoring forces in the same direction (towards the equilibrium position). Thus, their effective stiffnesses add directly: $K_{\text{eq}} = K_1 + K_2$ .

Answer

Displacing the mass by $x$ compresses one spring and elongates the other, giving a total restoring force of $F = -(K_1 + K_2)x$ , which yields an equivalent spring constant of $K_{\text{eq}} = K_1 + K_2$ and a time period of $T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .

Why each option works or fails

A: $\dfrac{1}{2\pi}\sqrt{\dfrac{\mathrm{K}_1 + \mathrm{K}_2}{\mathrm{m}}}$ - Confusing the formula for natural linear frequency $f = \frac{1}{2\pi}\sqrt{\frac{K_{\text{eq}}}{m}}$ with the time period $T = \frac{1}{f}$ . Remember that time period has units of time (seconds), so it must be proportional to $\sqrt{\frac{m}{K_{\text{eq}}}}$ , giving $T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .
B: $\dfrac{1}{2\pi}\sqrt{\dfrac{\mathrm{K}_1 - \mathrm{K}_2}{\mathrm{m}}}$ - Mistakenly subtracting the spring constants thinking the two opposing springs push against each other, while also inverting the formula to frequency instead of period. When the mass is displaced, both springs exert restoring forces in the same direction, so their spring constants add ( $K_{\text{eq}} = K_1 + K_2$ ), and the time period is $T = 2\pi\sqrt{m/K_{\text{eq}}}$ .
C · correct: $2\pi \sqrt{\dfrac{\mathrm{m}}{\mathrm{K}_1 + \mathrm{K}_2}}$ - None. This is the correct calculation. Both springs act in parallel because displacing the block by $x$ produces a combined restoring force $-(K_1 + K_2)x$ , giving $T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .
D: $2\pi \sqrt{\dfrac{\mathrm{m}}{\mathrm{K}_1 - \mathrm{K}_2}}$ - Assuming that because the springs are on opposite sides of the mass, their restoring forces oppose each other and subtract. A displacement to the right compresses the right spring and stretches the left spring; both resulting forces push and pull the mass to the left, adding together to give $K_{\text{eq}} = K_1 + K_2$ .

Step-by-step solution

given: Mass $m$ , spring constants $K_1$ and $K_2$ , frictionless surface.
find: Time period of oscillation $T$ of mass $m$ .
visualise: A mass $m$ is placed on a frictionless horizontal floor, attached to a fixed wall on the left via spring $K_1$ and a fixed wall on the right via spring $K_2$ . When displaced to the right by $x$ , the left spring is stretched by $x$ and pulls left with force $K_1 x$ ; the right spring is compressed by $2\pi \sqrt{\dfrac{\mathrm{m}}{\mathrm{K}_1 + \mathrm{K}_2}}$ and pushes left with force $K_2 x$ .
strategise: Find the net restoring force on the mass for a displacement $x$ : $F_{\text{net}} = -(K_1 + K_2)x$ . Relate this to $F = -K_{\text{eq}}x$ to find $K_{\text{eq}} = K_1 + K_2$ , then apply $T = 2\pi \sqrt{m/K_{\text{eq}}}$ .
execute: Acceleration $a = \frac{F_{\text{net}}}{m} = -\left(\frac{K_1 + K_2}{m}\right)x = -\omega^2 x$ , yielding $\omega = \sqrt{\frac{K_1 + K_2}{m}}$ . Thus, $K_2$ .
verify: Dimensionally $[T] = \text{s}$ . When $K_2 = 0$ , $T = 2\pi\sqrt{m/K_1}$ , which correctly matches a single spring-mass system.
given: Mass $m$ connected between two rigid walls with springs $K_1$ and $K_2$ .
find: Time period $T$ .
strategise: When a mass is sandwiched between two springs attached to fixed supports, both springs experience identical displacement $x$ . This is a parallel combination: $K_{\text{eq}} = K_1 + K_2$ .
execute: Substitute $K_{\text{eq}}$ into $T = 2\pi \sqrt{\frac{m}{K_{\text{eq}}}} = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$ .
verify: Directly matches Option (2).