Three simple harmonic motions in the same direction having the same amplitude and same period are superposed. If each differ in phase from the next by `45^(@)`, then

Options

A. the resultant amplitude is `(1+sqrt(2))`a
B. the phase of the resultant motion relative to the first is `tan^(-1)(1/2)`
C. the energy associated with the resulting motion is `(3+2sqrt(2))` times the energy associated with any single motion.
D. the resulting motion is not simple harmonic

Answer

To solve the problem of three simple harmonic motions (SHMs) that are superposed, we will follow these steps:

### Step 1: Define the SHMs
Let the three SHMs be defined as follows:
- \( X_1 = A \sin(\omega t) \)
- \( X_2 = A \sin(\omega t + 45^\circ) \)
- \( X_3 = A \sin(\omega t + 90^\circ) \)

### Step 2: Express the SHMs in Phasor Form
We can represent these SHMs as phasors:
- \( X_1 \) corresponds to a phasor at an angle of \( 0^\circ \)
- \( X_2 \) corresponds to a phasor at an angle of \( 45^\circ \)
- \( X_3 \) corresponds to a phasor at an angle of \( 90^\circ \)

### Step 3: Calculate the Resultant Amplitude
To find the resultant amplitude \( A_R \), we can use the principle of superposition. The resultant phasor can be calculated using vector addition:
- The phasor for \( X_1 \) is \( A \) at \( 0^\circ \).
- The phasor for \( X_2 \) is \( A \) at \( 45^\circ \).
- The phasor for \( X_3 \) is \( A \) at \( 90^\circ \).

Using the cosine and sine components:
- For \( X_1 \):
- \( X_{1x} = A \cos(0^\circ) = A \)
- \( X_{1y} = A \sin(0^\circ) = 0 \)

- For \( X_2 \):
- \( X_{2x} = A \cos(45^\circ) = \frac{A}{\sqrt{2}} \)
- \( X_{2y} = A \sin(45^\circ) = \frac{A}{\sqrt{2}} \)

- For \( X_3 \):
- \( X_{3x} = A \cos(90^\circ) = 0 \)
- \( X_{3y} = A \sin(90^\circ) = A \)

Now, summing the components:
- Total \( X \) component:
\[ X_R = A + \frac{A}{\sqrt{2}} + 0 = A + \frac{A}{\sqrt{2}} \]

- Total \( Y \) component:
\[ Y_R = 0 + \frac{A}{\sqrt{2}} + A = A + \frac{A}{\sqrt{2}} \]

### Step 4: Calculate the Resultant Amplitude
The resultant amplitude \( A_R \) is given by:
\[ A_R = \sqrt{(X_R)^2 + (Y_R)^2} \]
Substituting the values:
\[ A_R = \sqrt{\left(A + \frac{A}{\sqrt{2}}\right)^2 + \left(A + \frac{A}{\sqrt{2}}\right)^2} \]
\[ = \sqrt{2\left(A + \frac{A}{\sqrt{2}}\right)^2} = \sqrt{2}\left(A + \frac{A}{\sqrt{2}}\right) \]
\[ = \sqrt{2}\left(A(1 + \frac{1}{\sqrt{2}})\right) = (1 + \sqrt{2})A \]

### Step 5: Determine the Phase Constant
The phase constant \( \phi \) of the resultant motion can be determined using:
\[ \tan(\phi) = \frac{Y_R}{X_R} \]
Since both components are equal:
\[ \tan(\phi) = 1 \implies \phi = 45^\circ \]

### Step 6: Calculate the Energy of the Resultant Motion
The energy associated with a single SHM is given by:
\[ E = \frac{1}{2} k A^2 \]
For the resultant motion:
\[ E_R = \frac{1}{2} k A_R^2 = \frac{1}{2} k \left((1 + \sqrt{2})A\right)^2 \]
\[ = \frac{1}{2} k A^2 (1 + 2\sqrt{2} + 2) = \frac{1}{2} k A^2 (3 + 2\sqrt{2}) \]

### Conclusion
The resultant amplitude is \( (1 + \sqrt{2})A \), the phase constant is \( 45^\circ \), and the energy associated with the resultant motion is \( (3 + 2\sqrt{2}) \) times the energy associated with any single motion.