Question:
For a particle performing S.H.M., the displacement–time graph is as shown.
For a particle performing S.H.M., the displacement–time graph is as shown.
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In SHM, force and displacement are always $\pi$ radians (180°) out of phase.
Updated On: Jan 30, 2026
- (B)
- (C)
- (A)
- (D)
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The Correct Option is C
Solution and Explanation
Step 1: Force–displacement relation in SHM.
For simple harmonic motion:
\[ F = -kx \] Force is always proportional to displacement and opposite in direction.
Step 2: Relation between force and time.
Since displacement varies sinusoidally with time, force will also vary sinusoidally with time but will be opposite in phase to displacement.
Step 3: Phase relation.
When displacement is maximum positive, force is maximum negative and vice versa.
Step 4: Matching with given graphs.
Among the given options, graph (A) correctly represents a sinusoidal force–time graph that is opposite in phase to the displacement–time graph.
Step 5: Conclusion.
The correct force–time graph is option (A).
For simple harmonic motion:
\[ F = -kx \] Force is always proportional to displacement and opposite in direction.
Step 2: Relation between force and time.
Since displacement varies sinusoidally with time, force will also vary sinusoidally with time but will be opposite in phase to displacement.
Step 3: Phase relation.
When displacement is maximum positive, force is maximum negative and vice versa.
Step 4: Matching with given graphs.
Among the given options, graph (A) correctly represents a sinusoidal force–time graph that is opposite in phase to the displacement–time graph.
Step 5: Conclusion.
The correct force–time graph is option (A).
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