1. Condition for SHM

A motion is simple harmonic if it can be written as a single sinusoidal function of time:

$x(t)=A\sin(\Omega t+\phi) \quad \text{or} \quad x(t)=A\cos(\Omega t+\phi)$

Its period is

$T=\frac{2\pi}{\Omega}$

We need the function that represents SHM with period

$T=\frac{\pi}{\omega}$

So,

$\frac{2\pi}{\Omega}=\frac{\pi}{\omega}$

which gives

$\Omega=2\omega$

Thus the required function must be a single sine/cosine with angular frequency $2\omega$.

2. Check each option

Option A

$\cos(\omega t)+\cos(2\omega t)+\cos(3\omega t)$

This is a sum of three sinusoidal terms with different angular frequencies $\omega, 2\omega, 3\omega$.

So it is not a single harmonic function, hence not SHM.

Option B

The printed form is $\sin 2(\omega t)$, which is understood as

$\sin(2\omega t)$

This is a single sinusoidal function with angular frequency $2\omega$.

Hence it is SHM, and its period is

$T=\frac{2\pi}{2\omega}=\frac{\pi}{\omega}$

So Option B is correct.

Option C

$\sin(\omega t)+\cos(\omega t)$

Although this can be combined into a single sinusoid:

$\sin(\omega t)+\cos(\omega t)=\sqrt{2}\sin\left(\omega t+\frac{\pi}{4}\right)$

this has angular frequency $\omega$, so its period is

$T=\frac{2\pi}{\omega}$

which is not $\frac{\pi}{\omega}$.

So Option C is incorrect.

Option D

$3\cos\left(\frac{\pi}{4}-2\omega t\right)$

Using $\cos(-\theta)=\cos\theta$, this is equivalent to a cosine with angular frequency $2\omega$:

$3\cos\left(2\omega t-\frac{\pi}{4}\right)$

This is SHM with period

$T=\frac{2\pi}{2\omega}=\frac{\pi}{\omega}$

So Option D is also correct.

3. Conclusion

Both B and D represent simple harmonic motion with period $\dfrac{\pi}{\omega}$.

Therefore, as a single-correct MCQ, the question is flawed because it has two correct options.

The stored answer lists only D, but B is equally correct.