Question

A spring having a spring constant $ 1200{\text{N}}{{\text{m}}^{ - 1}} $ is mounted on a horizontal table. A mass of $ 3\;{\text{kg}} $ is attached to the free end of the spring. The mass is then pulled to a distance of $ 2.0\;{\text{cm}} $ and released. Determine maximum acceleration of mass.

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.
Formula Used: The formula to find out the maximum acceleration of the body is given by:
 $ a = {\omega ^2} \times A $
Where, $ a $ is the maximum acceleration of the body
 $ \omega $ is the angular frequency
 $ A $ is the maximum displacement.

Complete step by step answer:
According to the question,
The spring constant is given as $ k = 1200N/m $
The mass of the body is given as $ m = 3kg $
The displacement of the body sideways is given as $ A = 2cm = 0.02m $
The maximum acceleration of the body is given by the formula
 $ a = {\omega ^2} \times A $
We know that the angular frequency $ \omega = \sqrt {\dfrac{k}{m}} $
Where,
 $ k $ is the spring constant
 $ m $ is the mass of the body
Now we will substitute all the values in the above equation to get
 $ a = \dfrac{k}{m} \times A $
 $ \Rightarrow a = \dfrac{{1200}}{3} \times 0.02 $
Upon solving the equation, we get
 $ \therefore a = 8m/{s^2} $ .

Note:
Simple harmonic motion can be defined as an oscillatory motion in which the particle's acceleration at any position is directly proportional to the mean position's displacement. It is a special oscillatory motion case. 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 (SHM) being the most important.