Which of the following is/are not SHM?

Options

A. `y= A cos omega t`
B. `y = A sin omega t`
C. `y =A sin 3 omega t`
D. `y= A e^(kT)`

Answer

To determine which of the given equations does not represent Simple Harmonic Motion (SHM), we need to analyze each equation based on the condition that the acceleration \( a \) must be directly proportional to the negative of the displacement \( y \) from the mean position. Mathematically, this can be expressed as:

\[ a = -k y \]

where \( k \) is a positive constant.

Let's analyze each option step by step:

### Step 1: Analyze the first equation \( y = A \cos(\omega t) \)

1. **Calculate Velocity**:
\[ v = \frac{dy}{dt} = -A \omega \sin(\omega t) \]

2. **Calculate Acceleration**:
\[ a = \frac{dv}{dt} = -A \omega^2 \cos(\omega t) \]

3. **Relate Acceleration to Displacement**:
\[ a = -\omega^2 y \]
This shows that acceleration is directly proportional to \(-y\). Thus, this equation represents SHM.

### Step 2: Analyze the second equation \( y = A \sin(\omega t) \)

1. **Calculate Velocity**:
\[ v = \frac{dy}{dt} = A \omega \cos(\omega t) \]

2. **Calculate Acceleration**:
\[ a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \]

3. **Relate Acceleration to Displacement**:
\[ a = -\omega^2 y \]
This also shows that acceleration is directly proportional to \(-y\). Thus, this equation represents SHM.

### Step 3: Analyze the third equation \( y = A \sin(3\omega t) \)

1. **Calculate Velocity**:
\[ v = \frac{dy}{dt} = 3A \omega \cos(3\omega t) \]

2. **Calculate Acceleration**:
\[ a = \frac{dv}{dt} = -9A \omega^2 \sin(3\omega t) \]

3. **Relate Acceleration to Displacement**:
\[ a = -9\omega^2 y \]
This shows that acceleration is directly proportional to \(-y\). Thus, this equation represents SHM.

### Step 4: Analyze the fourth equation \( y = e^{kt} \)

1. **Calculate Velocity**:
\[ v = \frac{dy}{dt} = k e^{kt} \]

2. **Calculate Acceleration**:
\[ a = \frac{dv}{dt} = k^2 e^{kt} \]

3. **Relate Acceleration to Displacement**:
Here, we see that:
\[ a = k^2 y \]
This indicates that acceleration is directly proportional to \(y\), not \(-y\). Thus, this equation does not represent SHM.

### Conclusion:
The fourth equation \( y = e^{kt} \) does not satisfy the condition for SHM, as its acceleration is not proportional to the negative of its displacement.

### Final Answer:
The equation that is **not SHM** is:
\[ y = e^{kt} \]

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