Question

A particle moves according to the equation x = Asin(\(\omega t\)). The potential energy is maximum at time \(t = \frac{T}{2\beta}\), where T is the time period of particle. Find the minimum value of \(\beta\) :

Hint:

Remember the standard SHM timeline: at \(t=0\), \(x=0\) (min PE); at \(t=T/4\), \(x=A\) (max PE); at \(t=T/2\), \(x=0\) (min PE); at \(t=3T/4\), \(x=-A\) (max PE).

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), the potential energy (\(U\)) is given by \(U = \frac{1}{2} k x^2\).
Potential energy is maximum when the displacement \(x\) is maximum, i.e., \(x = \pm A\).
Step 2: Key Formula or Approach:
1. Displacement \(x = A \sin(\omega t)\).
2. Maximum displacement occurs when \(\sin(\omega t) = \pm 1\).
3. Relationship between angular frequency and time period: \(\omega = \frac{2\pi}{T}\).
Step 3: Detailed Explanation:
For \(U\) to be maximum, we need:
\[ \sin(\omega t) = 1 \]
The first time this occurs is when:
\[ \omega t = \frac{\pi}{2} \]
Substituting \(\omega = \frac{2\pi}{T}\):
\[ \left( \frac{2\pi}{T} \right) t = \frac{\pi}{2} \]
\[ t = \frac{T}{4} \]
We are given that this time is \(t = \frac{T}{2\beta}\). Equating the two expressions:
\[ \frac{T}{4} = \frac{T}{2\beta} \]
\[ 2\beta = 4 \implies \beta = 2 \]
Step 4: Final Answer:
The minimum value of \(\beta\) is 2.