A particle is executing SHM with total mechanical energy 90J and amplitude 6cm. If its energy is somehow decreased to 40J then its amplitude will become

Options

A. 2cm
B. 4cm
C. `(8)/(3)cm`
D. `(4)/(3)cm`

Answer

To solve the problem step by step, we will use the relationship between the total mechanical energy in simple harmonic motion (SHM) and the amplitude.

### Step 1: Understanding the relationship between energy and amplitude
The total mechanical energy (E) in SHM is given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant and \( A \) is the amplitude.

### Step 2: Establishing the initial conditions
From the problem, we know:
- Initial energy \( E_1 = 90 \, \text{J} \)
- Initial amplitude \( A_1 = 6 \, \text{cm} = 0.06 \, \text{m} \)

### Step 3: Finding the spring constant (k)
Using the formula for total energy:
\[ E_1 = \frac{1}{2} k A_1^2 \]
we can rearrange it to find \( k \):
\[ k = \frac{2E_1}{A_1^2} \]
Substituting the values:
\[ k = \frac{2 \times 90}{(0.06)^2} \]
\[ k = \frac{180}{0.0036} \]
\[ k = 50000 \, \text{N/m} \]

### Step 4: Establishing the new conditions
Now, the energy is decreased to:
- Final energy \( E_2 = 40 \, \text{J} \)

### Step 5: Finding the new amplitude (A2)
Using the same energy formula for the final conditions:
\[ E_2 = \frac{1}{2} k A_2^2 \]
Rearranging gives:
\[ A_2^2 = \frac{2E_2}{k} \]
Substituting the known values:
\[ A_2^2 = \frac{2 \times 40}{50000} \]
\[ A_2^2 = \frac{80}{50000} \]
\[ A_2^2 = 0.0016 \]
Taking the square root to find \( A_2 \):
\[ A_2 = \sqrt{0.0016} \]
\[ A_2 = 0.04 \, \text{m} = 4 \, \text{cm} \]

### Final Answer
The new amplitude when the energy is decreased to 40 J is:
\[ A_2 = 4 \, \text{cm} \]

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