The diagram shows two oscillations. What is the phase difference between the oscillations?

The acceleration of a particle starting from rest varies with time according to the relation $A = -a\omega^2 \sin\omega t$. The displacement of this particle at a time $t$ will be

The phase difference between two $SHM$ equations $y_1 = 10 \sin(10\pi t + \frac{\pi}{3})$ and $y_2 = 12 \sin(8\pi t + \frac{\pi}{4})$ at $t = 0.5 \ s$ is:

$A$ particle executing a simple harmonic motion has a period of $6 \,s$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

Explain with plots the position of a particle executing simple harmonic motion at different times.

$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.