A mass `m=100 ` gram is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equalt to `0.16m` and time period equal to 2 second. Initially the mass is released from rest at `t=0` and displacement `x=-0.16` metre. The expression for the displacement of mass at any time `(t)` is

Options

A. (a)`x=0.16cos(pit)`
B. (b)`x=-016cos(pit)`
C. (c)`x=0.16sin(pit+pi)`
D. (d)`x=-0.16sin(pit+pi)`

Explanation

To derive the expression for the displacement of the mass attached to the spring, we will follow these steps:

### Step 1: Identify the parameters
Given:
- Mass \( m = 100 \) grams (which we convert to kilograms: \( m = 0.1 \) kg)
- Amplitude \( A = 0.16 \) m
- Time period \( T = 2 \) seconds

### Step 2: Determine the angular frequency
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{2} = \pi \text{ rad/s} \]

### Step 3: Write the general equation for displacement
The general equation for the displacement \( x(t) \) in simple harmonic motion (SHM) is given by:
\[ x(t) = A \cos(\omega t + \phi) \]
where \( \phi \) is the phase constant.

### Step 4: Determine the phase constant
Since the mass is released from rest at \( t = 0 \) with a displacement of \( x = -0.16 \) m, we can substitute these values into the equation to find \( \phi \):
\[ -0.16 = 0.16 \cos(\pi \cdot 0 + \phi) \]
This simplifies to:
\[ -1 = \cos(\phi) \]
Thus, \( \phi = \pi \) (since the cosine of \( \pi \) is -1).

### Step 5: Substitute values into the displacement equation
Now we can substitute \( A \), \( \omega \), and \( \phi \) back into the displacement equation:
\[ x(t) = 0.16 \cos(\pi t + \pi) \]
Using the property of cosine, \( \cos(\theta + \pi) = -\cos(\theta) \):
\[ x(t) = 0.16 \cos(\pi t + \pi) = -0.16 \cos(\pi t) \]

### Final Expression
Thus, the expression for the displacement of the mass at any time \( t \) is:
\[ x(t) = -0.16 \cos(\pi t) \]