Question:
If the amplitude of linear S.H.M. is decreased then
If the amplitude of linear S.H.M. is decreased then
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Remember that the absolute definition of harmonicity means that the frequency and period are amplitude-independent properties (isochronism). Thus, you can immediately eliminate options (A), (B), and (C) during an exam without doing any math, as they all incorrectly predict a change in the period!
Updated On: Jun 18, 2026
- its period and total energy will increase.
- its period will increase and total energy will decrease.
- its period and total energy will decrease.
- its period will not change but total energy will decrease.
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The question asks about the effects on the periodic time ($T$) and the total mechanical energy ($E$) of an object performing linear simple harmonic motion when its maximum displacement amplitude ($A$) is reduced.
Step 2: Key Formula or Approach:
1. The time period ($T$) of a linear simple harmonic oscillator depends purely on the mass ($m$) and force constant ($k$): $$T = 2\pi\sqrt{\frac{m}{k}}$$ 2. The total mechanical energy ($E$) of a simple harmonic oscillator is given by: $$E = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} k A^2$$ This reveals that $T$ is entirely independent of amplitude ($T \propto A^0$), while total energy scales quadratically with amplitude ($E \propto A^2$).
Step 3: Detailed Explanation:
Let's analyze the properties based on our formulas: Time Period ($T$): Since the formula $T = 2\pi\sqrt{\frac{m}{k}}$ contains no amplitude parameter $A$, changing the amplitude has absolutely zero impact on the time it takes to complete one oscillation cycle. Thus, the period will not change. Total Energy ($E$): The energy relationship shows $E \propto A^2$. If the amplitude is decreased, its square also decreases, which causes the total mechanical energy of the system to decrease. Combining these two conclusions matches description option (D).
Step 4: Final Answer:
The periodic time remains unchanged, but the total energy decreases, corresponding to option (D).
The question asks about the effects on the periodic time ($T$) and the total mechanical energy ($E$) of an object performing linear simple harmonic motion when its maximum displacement amplitude ($A$) is reduced.
Step 2: Key Formula or Approach:
1. The time period ($T$) of a linear simple harmonic oscillator depends purely on the mass ($m$) and force constant ($k$): $$T = 2\pi\sqrt{\frac{m}{k}}$$ 2. The total mechanical energy ($E$) of a simple harmonic oscillator is given by: $$E = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} k A^2$$ This reveals that $T$ is entirely independent of amplitude ($T \propto A^0$), while total energy scales quadratically with amplitude ($E \propto A^2$).
Step 3: Detailed Explanation:
Let's analyze the properties based on our formulas: Time Period ($T$): Since the formula $T = 2\pi\sqrt{\frac{m}{k}}$ contains no amplitude parameter $A$, changing the amplitude has absolutely zero impact on the time it takes to complete one oscillation cycle. Thus, the period will not change. Total Energy ($E$): The energy relationship shows $E \propto A^2$. If the amplitude is decreased, its square also decreases, which causes the total mechanical energy of the system to decrease. Combining these two conclusions matches description option (D).
Step 4: Final Answer:
The periodic time remains unchanged, but the total energy decreases, corresponding to option (D).
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