Out of the following functions representing motion of a particle which represents SHM?
1. `x=sin^(3)omegat`
2. `x=1+omegat+omega^(2)t^(2)`
3. `x=cosomegat+cos3omegat+cos5omegat`
4. `x=sinomegat+cosomegat`

Options

A. `y = sin omega t - cos omega t`
B. `y = sin^(3) omega t`
C. `y = 5 cos ((3pi)/(4) - 3omega t)`
D. `y=1 + omega t + omega^(2) t^(2)`

Answer

To determine which of the given functions represents Simple Harmonic Motion (SHM), we need to analyze each function based on the criteria for SHM. The key characteristic of SHM is that the acceleration (A) of the particle is directly proportional to the negative of its displacement (x), which can be mathematically expressed as:

\[ A = -k x \]

where \( k \) is a positive constant.

Now, let's evaluate each of the given functions:

1. **Function 1: \( x = \sin^3(\omega t) \)**
To check if this represents SHM, we need to find the second derivative with respect to time (t).
- First, differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = 3\sin^2(\omega t)\cos(\omega t) \cdot \omega \]
- Now, differentiate \( v \) to find acceleration \( A \):
\[ A = \frac{dv}{dt} = 3\left(2\sin(\omega t)\cos(\omega t)\cdot\omega^2\cos(\omega t) - \sin^2(\omega t)\cdot\omega^2\sin(\omega t)\right) \]
- The resulting expression does not have a linear relationship with \( x \), hence it does not represent SHM.

2. **Function 2: \( x = 1 + \omega t + \omega^2 t^2 \)**
This is a quadratic function of time.
- The first derivative (velocity) will be:
\[ v = \frac{dx}{dt} = \omega + 2\omega^2 t \]
- The second derivative (acceleration) will be:
\[ A = \frac{dv}{dt} = 2\omega^2 \]
- Since acceleration is constant and does not depend on displacement \( x \), this function does not represent SHM.

3. **Function 3: \( x = \cos(\omega t) + \cos(3\omega t) + \cos(5\omega t) \)**
This function is a sum of cosine functions with different frequencies.
- To check for SHM, we can differentiate:
\[ v = \frac{dx}{dt} = -\omega \sin(\omega t) - 3\omega \sin(3\omega t) - 5\omega \sin(5\omega t) \]
- Differentiating again gives:
\[ A = \frac{dv}{dt} = -\omega^2 \cos(\omega t) - 9\omega^2 \cos(3\omega t) - 25\omega^2 \cos(5\omega t) \]
- The acceleration is not proportional to \( -x \) as it involves multiple frequencies, hence it does not represent SHM.

4. **Function 4: \( x = \sin(\omega t) + \cos(\omega t) \)**
This function can be simplified using the identity for sine and cosine.
- First, differentiate:
\[ v = \frac{dx}{dt} = \omega \cos(\omega t) - \omega \sin(\omega t) \]
- Differentiate again:
\[ A = \frac{dv}{dt} = -\omega^2 \sin(\omega t) - \omega^2 \cos(\omega t) = -\omega^2(\sin(\omega t) + \cos(\omega t)) \]
- Here, we see that acceleration is proportional to \( -x \), satisfying the condition for SHM.

**Conclusion:** The function that represents Simple Harmonic Motion is **Option 4: \( x = \sin(\omega t) + \cos(\omega t) \)**.