Question

A particle executes SHM with time period T. If acceleration is doubled keeping amplitudes constant, new time period is

• A. T
• B. $\frac{T}{2}$
• C. 2T
• D. $\frac{T}{\sqrt{2}}$
• E. $\sqrt{2} T$

Hint:

Time period is inversely proportional to angular frequency ($T \propto 1/\omega$). Max acceleration is proportional to the square of angular frequency ($a \propto \omega^2$). Thus, $a \propto 1/T^2$, or $T \propto 1/\sqrt{a}$. If $a$ is doubled ($2\times$), $T$ becomes $1/\sqrt{2}$ times.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), the acceleration of a particle is directly proportional to its displacement but in the opposite direction. The maximum acceleration occurs at the extreme positions (maximum displacement, which is the amplitude). The relationships between time period, angular frequency, acceleration, and amplitude dictate the dynamics of the system.
Step 2: Key Formula or Approach:
1. Maximum acceleration: \(a_{max} = \omega^2 A\)
2. Time period: \(T = \frac{2\pi}{\omega}\)
Where \(\omega\) is angular frequency and \(A\) is amplitude.
We will find the new angular frequency \(\omega'\) when \(a_{max}\) is doubled, and then find the new time period \(T'\).
Step 3: Detailed Explanation:
Let the initial parameters be \(T\), \(\omega\), \(A\), and \(a_{max}\).
\[ a_{max} = \omega^2 A \]
\[ T = \frac{2\pi}{\omega} \]
The problem states the new acceleration is doubled while keeping amplitude constant. This implies the maximum acceleration is doubled.
Let the new parameters be \(T'\), \(\omega'\), \(A'\), and \(a'_{max}\).
Given:
\(A' = A\) (amplitude is constant)
\(a'_{max} = 2 a_{max}\)
Using the acceleration formula for the new state:
\[ a'_{max} = (\omega')^2 A' \]
Substitute the given relationships:
\[ 2 a_{max} = (\omega')^2 A \]
Since we know \(a_{max} = \omega^2 A\), substitute this in:
\[ 2 (\omega^2 A) = (\omega')^2 A \]
Cancel \(A\) from both sides (since \(A \neq 0\)):
\[ 2\omega^2 = (\omega')^2 \]
Take the square root:
\[ \omega' = \sqrt{2}\omega \]
Now, find the new time period \(T'\):
\[ T' = \frac{2\pi}{\omega'} \]
Substitute \(\omega' = \sqrt{2}\omega\):
\[ T' = \frac{2\pi}{\sqrt{2}\omega} \]
We know the initial time period is \(T = \frac{2\pi}{\omega}\), so we can substitute \(T\):
\[ T' = \frac{T}{\sqrt{2}} \]
Step 4: Final Answer:
The new time period is \(\frac{T}{\sqrt{2}}\).