A horizontal spring mass system is executing SHM with time period of 4s. At time t=0, it is at mean position. Find the minimum time after which its potential energy becomes three times of kinetic energy.

Options

A. 1s
B. `1/2`s
C. `1/3`s
D. `2/3`s

Answer

To solve the problem step by step, we will analyze the motion of the spring-mass system executing Simple Harmonic Motion (SHM) and derive the necessary equations for potential energy (PE) and kinetic energy (KE). We will then find the time when the potential energy becomes three times the kinetic energy.

### Step 1: Understand the SHM parameters
The time period \( T \) of the SHM is given as 4 seconds. At time \( t = 0 \), the mass is at the mean position, which means the displacement \( y = 0 \).

### Step 2: Write the equations for SHM
The displacement in SHM can be expressed as:
\[ y(t) = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency. Since the mass is at the mean position at \( t = 0 \), we can set the phase constant \( \phi = 0 \).

### Step 3: Calculate the angular frequency
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting \( T = 4 \) seconds:
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{rad/s} \]

### Step 4: Write the expressions for potential energy and kinetic energy
The potential energy \( PE \) in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
Using \( k = m \omega^2 \) and \( x = A \sin(\omega t) \):
\[ PE = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t) \]

The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( v = \frac{dy}{dt} = A \omega \cos(\omega t) \):
\[ KE = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]

### Step 5: Set up the equation for the condition given
We want to find the time \( t \) when the potential energy is three times the kinetic energy:
\[ PE = 3 \times KE \]
Substituting the expressions for \( PE \) and \( KE \):
\[ \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t) = 3 \left( \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \right) \]
Cancelling common terms:
\[ \sin^2(\omega t) = 3 \cos^2(\omega t) \]

### Step 6: Use the identity to relate sine and cosine
Using the identity \( \tan^2(\omega t) = \frac{\sin^2(\omega t)}{\cos^2(\omega t)} \):
\[ \tan^2(\omega t) = 3 \]
Taking the square root:
\[ \tan(\omega t) = \sqrt{3} \]

### Step 7: Find the angle corresponding to the tangent value
The angle \( \omega t \) where \( \tan(\omega t) = \sqrt{3} \) is:
\[ \omega t = \frac{\pi}{3} \]

### Step 8: Solve for time \( t \)
Substituting \( \omega = \frac{\pi}{2} \):
\[ \frac{\pi}{2} t = \frac{\pi}{3} \]
Solving for \( t \):
\[ t = \frac{2}{3} \, \text{seconds} \]

### Final Answer
The minimum time after which the potential energy becomes three times the kinetic energy is:
\[ t = \frac{2}{3} \, \text{seconds} \]