Question 318: Oscillations

Question

If the amplitude of linear S.H.M. is decreased then

Accepted Answer

To answer this question, it's essential to understand the characteristics of Simple Harmonic Motion (S.H.M.) and the effects of amplitude on its various aspects.
Period of S.H.M.:
The period of a simple harmonic oscillator is the time it takes for one complete cycle of motion. For a mass-spring system, the period $T$ is given by:
$$ T = 2\pi \sqrt{\frac{m}{k}} $$
where $m$ is the mass of the object and $k$ is the spring constant. Notice that the period $T$ depends only on the mass and the spring constant, and not on the amplitude $A$ of the oscillation. Thus, decreasing the amplitude does not affect the period of the motion.
Total Energy of S.H.M.:
The total energy $E$ in simple harmonic motion is proportional to the square of its amplitude $A$. For a mass-spring system, the total energy is given by:
$$ E = \frac{1}{2} k A^2 $$
Given this relationship, if the amplitude is decreased, the total energy will also decrease.
Therefore, based on this analysis:
Option A
: Incorrect. The period does not increase with a decrease in amplitude, and the total energy decreases rather than increases.
Option B
: Incorrect. The period does not increase with a decrease in amplitude, though the total energy does decrease.
Option C
: Incorrect. The period does not decrease with a decrease in amplitude, though the total energy does decrease.
Option D
: Correct. The period remains unchanged, but the total energy decreases with a decrease in amplitude.
Hence, the correct answer is:
Option D:
its period will not change but total energy will decrease
.

Answer

its period and total energy will increase.

Answer

its period will increase and total energy will decrease

Answer

its period and total energy will decrease.

Answer

its period will not change but total energy will decrease

Oscillations

If the amplitude of linear S.H.M. is decreased then

Choose the correct answer: