Springs In Combination
Two identical springs each of force constant k are connected in parallel and support a mass m, so how does the oscillation period compare to a single spring?
Select the correct option:
Solution
It decreases by a factor of the square root of two
Springs combined in parallel share the load so their stiffnesses add, a result following from the force analysis in NCERT Class 11, Chapter 14 (Oscillations); the effective constant becomes keff=k+k=2k. Since the period of a spring-mass system is T=2πm/keff, a larger effective stiffness gives a shorter period. Comparing the parallel case with a single spring, TsingleTparallel=2kk=21. Thus the period decreases by a factor of 2. The option that it increases by 2 describes a series combination, where the effective constant is halved, not doubled. The option that it stays the same ignores that parallel springs stiffen the system. The option that it halves would require the effective constant to quadruple, which is not the case for two equal springs. The physical reason the constants add in parallel is that both springs stretch by the same displacement x, so each contributes a restoring force kx and the total is 2kx, equivalent to a single spring of constant 2k. Contrast this with a series arrangement, where the same force stretches both springs and the extensions add, giving a softer effective constant k/2 and a longer period. A sanity check: stiffer support should make oscillations faster, so a smaller period is physically expected, and the factor 1/2≈0.71 is a modest, reasonable reduction rather than an extreme change.
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About This Question
- Subject
- physics
- Chapter
- oscillations and waves
- Topic
- springs in combination
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
It decreases by a factor of the square root of two
Springs combined in parallel share the load so their stiffnesses add, a result following from the force analysis in NCERT Class 11, Chapter 14 (Oscillations); the effective constant becomes keff=k+k=2k. Since the period of a spring-mass system is T=2πm/keff, a larger effective stiffness gives a shorter period. Comparing the parallel case with a single spring, TsingleTparallel=2kk=21. Thus the period decreases by a factor of 2. The option that it increases by 2 describes a series combination, where the effective constant is halved, not doubled. The option that it stays the same ignores that parallel springs stiffen the system. The option that it halves would require the effective constant to quadruple, which is not the case for two equal springs. The physical reason the constants add in parallel is that both springs stretch by the same displacement x, so each contributes a restoring force kx and the total is 2kx, equivalent to a single spring of constant 2k. Contrast this with a series arrangement, where the same force stretches both springs and the extensions add, giving a softer effective constant k/2 and a longer period. A sanity check: stiffer support should make oscillations faster, so a smaller period is physically expected, and the factor 1/2≈0.71 is a modest, reasonable reduction rather than an extreme change.
This medium difficulty physics question is from the chapter oscillations and waves, covering the topic of springs in combination. It appeared in the 2025 exam.
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