A body of mass 100 g attached to a spring executed SHM of period 2 s and amplitude 10 cm. How long a time is required for it to move from a point 5 cm below its equilibrium position to a point 5 cm above it, when it makes simple harmonic vertical oscillation (take `g=10(m)/(s^2)`)?

Options

A. 0.6 s
B. `(1)/(3)s`
C. `1.5s`
D. `2.2s`

Explanation

To solve the problem, we need to determine the time taken for a body executing simple harmonic motion (SHM) to move from a point 5 cm below its equilibrium position to a point 5 cm above it. Here’s a step-by-step solution:

### Step 1: Understand the parameters of the SHM
- Mass of the body, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Period, \( T = 2 \, \text{s} \)
- Amplitude, \( A = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Step 2: Determine the angular frequency \( \omega \)
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: Set up the displacement equation
The displacement \( x \) in SHM can be expressed as:
\[ x(t) = A \cos(\omega t) \]
For our case:
\[ x(t) = 0.1 \cos(\pi t) \]

### Step 4: Calculate the time \( t_1 \) to reach 5 cm below the equilibrium position
At 5 cm below the equilibrium position:
\[ x = -5 \, \text{cm} = -0.05 \, \text{m} \]
Setting up the equation:
\[ -0.05 = 0.1 \cos(\pi t_1) \]
This simplifies to:
\[ -0.5 = \cos(\pi t_1) \]
Taking the inverse cosine:
\[ \pi t_1 = \cos^{-1}(-0.5) \]
The angle corresponding to \( \cos^{-1}(-0.5) \) is \( \frac{2\pi}{3} \) (in the second quadrant). Therefore:
\[ t_1 = \frac{2}{3} \, \text{s} \]

### Step 5: Calculate the time \( t_2 \) to reach 5 cm above the equilibrium position
At 5 cm above the equilibrium position:
\[ x = 5 \, \text{cm} = 0.05 \, \text{m} \]
Setting up the equation:
\[ 0.05 = 0.1 \cos(\pi t_2) \]
This simplifies to:
\[ 0.5 = \cos(\pi t_2) \]
Taking the inverse cosine:
\[ \pi t_2 = \cos^{-1}(0.5) \]
The angle corresponding to \( \cos^{-1}(0.5) \) is \( \frac{\pi}{3} \) (in the first quadrant). Therefore:
\[ t_2 = \frac{1}{3} \, \text{s} \]

### Step 6: Calculate the time taken to move from point P to point Q
The total time taken to move from point P (5 cm below) to point Q (5 cm above) is:
\[ \Delta t = t_2 - t_1 = \frac{1}{3} - \frac{2}{3} = \frac{1}{3} \, \text{s} \]

### Final Answer
The time required for the body to move from a point 5 cm below its equilibrium position to a point 5 cm above it is \( \frac{1}{3} \) seconds.

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