If the displacement (y in m) and velocity (v in `ms^(-1))` of a particle executing SHM are related by the equation `4v^(2)=16-y^(2)`, then the path length of the motion and time period of oscillation respectively are

Options

A. `4 m, 2pi s`
B. `4 m, 4pi s`
C. `8 m, 2pi s`
D. `8 m, 4pi s`

Explanation

To solve the problem, we need to find the path length of the motion and the time period of oscillation for a particle executing simple harmonic motion (SHM) given the relationship between displacement \( y \) and velocity \( v \) as:

\[ 4v^2 = 16 - y^2 \]

### Step 1: Rearranging the equation
First, we can rearrange the given equation to express \( v \) in terms of \( y \):

\[ 4v^2 = 16 - y^2 \]

Dividing both sides by 4:

\[ v^2 = 4 - \frac{y^2}{4} \]

Taking the square root gives:

\[ v = \sqrt{4 - \frac{y^2}{4}} = \frac{1}{2} \sqrt{16 - y^2} \]

### Step 2: Relating velocity to SHM parameters
In SHM, the velocity \( v \) can also be expressed in terms of angular frequency \( \omega \) and amplitude \( a \):

\[ v = \omega \sqrt{a^2 - y^2} \]

### Step 3: Comparing the two expressions for velocity
Now we can compare the two expressions for \( v \):

\[ \frac{1}{2} \sqrt{16 - y^2} = \omega \sqrt{a^2 - y^2} \]

From the rearranged equation, we can identify:

1. \( \omega = \frac{1}{2} \)
2. \( a^2 = 16 \) which gives \( a = 4 \) (since amplitude is positive).

### Step 4: Finding the time period
The time period \( T \) of SHM is given by the formula:

\[ T = \frac{2\pi}{\omega} \]

Substituting \( \omega = \frac{1}{2} \):

\[ T = \frac{2\pi}{\frac{1}{2}} = 4\pi \text{ seconds} \]

### Step 5: Finding the path length
The path length \( L \) in SHM is given by:

\[ L = 2a \]

Substituting \( a = 4 \):

\[ L = 2 \times 4 = 8 \text{ meters} \]

### Final Answer
Thus, the path length of the motion is \( 8 \) meters and the time period of oscillation is \( 4\pi \) seconds.

### Summary of Results
- Path Length: \( 8 \) meters
- Time Period: \( 4\pi \) seconds

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