A mass of 2kg is attached to the spring of spring constant `50Nm^(-1)`. The block is pulled to a distance of 5 cm from its equilibrium position at `x=0` on a horizontal frictionless surface from rest at t=0. Write the expression for its displacement at anytime t.

Options

A. x=0.05sin5t m
B. x=0.05cos5t m
C. x=0.5sin5t m
D. x=5sin5t m

Explanation

To find the expression for the displacement of a mass attached to a spring at any time \( t \), we can follow these steps:

### Step 1: Identify the parameters
We have:
- Mass \( m = 2 \, \text{kg} \)
- Spring constant \( k = 50 \, \text{N/m} \)
- Initial displacement (amplitude) \( A = 5 \, \text{cm} = 0.05 \, \text{m} \)

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ \omega = \sqrt{\frac{50 \, \text{N/m}}{2 \, \text{kg}}} = \sqrt{25} = 5 \, \text{rad/s} \]

### Step 3: Determine the form of the displacement equation
The general form of the displacement \( x(t) \) for simple harmonic motion can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( \phi \) is the phase constant.

### Step 4: Determine the phase constant \( \phi \)
Since the block is pulled to a distance of 5 cm from its equilibrium position at \( t = 0 \) and released from rest, we can start from the equilibrium position. At \( t = 0 \), the displacement \( x(0) = 0 \). This condition is satisfied by using the sine function (as sine of zero is zero). Therefore, we can set \( \phi = 0 \).

### Step 5: Write the final expression for displacement
Substituting the values of \( A \) and \( \omega \) into the displacement equation:
\[ x(t) = 0.05 \sin(5t) \]

### Final Answer
The expression for the displacement at any time \( t \) is:
\[ x(t) = 0.05 \sin(5t) \]

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