Two simple harmonic motions `y_(1) = Asinomegat` and `y_(2)` = Acos`omega`t are superimposed on a particle of mass m. The total mechanical energy of the particle is

Options

A. `1/2momega^(2)A^(2)`
B. `momega^(2)A^(2)`
C. `1/4momega^(2)A^(2)`
D. zero

Answer

To find the total mechanical energy of a particle undergoing two superimposed simple harmonic motions (SHMs) given by \( y_1 = A \sin(\omega t) \) and \( y_2 = A \cos(\omega t) \), we can follow these steps:

### Step 1: Understand the Superposition of SHMs
The two SHMs can be combined to find the resultant displacement:
\[ y = y_1 + y_2 = A \sin(\omega t) + A \cos(\omega t) \]

### Step 2: Rewrite the Cosine Function
We can express \( \cos(\omega t) \) in terms of sine:
\[ \cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \]
Thus, we can rewrite the total displacement as:
\[ y = A \sin(\omega t) + A \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Step 3: Use Phasor Representation
In phasor representation, the two components \( y_1 \) and \( y_2 \) can be represented as vectors:
- \( y_1 \) makes an angle of \( 0^\circ \) with the x-axis.
- \( y_2 \) makes an angle of \( 90^\circ \) with the x-axis.

### Step 4: Calculate the Resultant Amplitude
Using the Pythagorean theorem (since the phase difference is \( 90^\circ \)):
\[ A_r = \sqrt{A^2 + A^2} = \sqrt{2A^2} = A\sqrt{2} \]

### Step 5: Determine the Total Mechanical Energy
The total mechanical energy \( E \) of a simple harmonic oscillator is given by:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant. We can express \( k \) in terms of \( \omega \) and \( m \):
\[ k = m \omega^2 \]
Substituting \( k \) into the energy formula:
\[ E = \frac{1}{2} (m \omega^2) (A_r^2) \]
Substituting \( A_r = A\sqrt{2} \):
\[ E = \frac{1}{2} (m \omega^2) (A\sqrt{2})^2 = \frac{1}{2} (m \omega^2) (2A^2) = m \omega^2 A^2 \]

### Final Answer
Thus, the total mechanical energy of the particle is:
\[ E = m \omega^2 A^2 \]