Question

A particle of mass 100g is executing SHM with amplitude of $10\;{\text{cm}}$. When the particle passes through the mean position at ${\text{t}} = 0$ its kinetic energy is $8\;{\text{mJ}}$. What is the equation of simple harmonic motion if the initial phase is zero?
(A) $y = 0.1\cos 4t$
(B) $y = 0.1\sin 4t$
(C) $y = 0.1\cos 2t$
(D) $y = 0.1\sin 2t$

Answer 1

Hint: Simple Harmonic Motion or SHM is defined as a motion in which the restored force from its mean position is directly proportional to the body's displacement. This restoring force's direction is always toward the mean position. This question can easily be solved by substituting the different values provided in the question into the equation of motion.
Formula Used: The displacement x of a body in motion is given by the formula
\[X = A\sin (\omega t + \phi )\]
Where
\[X\] is the displacement at a particular time
\[t\] is the time

Complete step by step answer:
The following quantities are given according to the question
The mass of the particle, \[m = 0.1kg\]
The amplitude of the particle, \[A = 0.1m\]
For any particle which passes through the mean position, the value of the kinetic energy will be maximum that is \[K{E_{\max }} = 8 \times {10^{ - 3}}J\]
Also, for any particle which passes through the mean position, the velocity of that particle will be maximum, that is \[{V_{\max }} = A\omega \]
The initial phase of oscillation, \[\phi = {0^ \circ }\]
 The equation of motion is given by \[X = A\sin (\omega t + \phi )\]
Now we know that the kinetic energy,
\[K{E_{\max }} = \dfrac{1}{2}m{v^2}_{\max }\]
\[ \Rightarrow K{E_{\max }} = 8 \times {10^{ - 3}}J\]
Now we will substitute the values from the given section into the above equation of kinetic energy
So, we get
\[ \Rightarrow \dfrac{1}{2} \times 0.1 \times {A^2}{\omega ^2} = 8 \times {10^{ - 3}}\]
\[ \Rightarrow 0.1 \times 0.1{\omega ^2} = 16 \times {10^{ - 3}}\]
On solving further, we get
\[\therefore \omega = 4rad/\sec \]
Now we will put the above value of \[\omega \] in the equation of motion. So, we get
\[X = A\sin (\omega t + \phi )\]
\[X = 0.1\sin (4t + {0^ \circ })\]
So, we get our desired result
\[\therefore X = 0.1\sin (4t)\]

Hence, the correct option is (B.)

Note: All simple harmonic movements are oscillatory, and SHM is also periodic, but not all oscillatory movements. Oscillatory motion is also referred to as the harmonic motion of all oscillatory movements, with simple harmonic motion being the most important (SHM).