Question

A 10 kg metal block is attached to a spring of spring constant $1000 N{m^{ - 1}}$. A block is displaced from equilibrium position by 10 cm and released. The maximum acceleration of the block is
(A) $10 m{s^{ - 2}}$
(B) $100 m{s^{ - 2}}$
(C) $200 m{s^{ - 2}}$
(D) $0.1 m{s^{ - 2}}$

Answer 1

In this question, we need to determine the maximum acceleration of the block such that the block is displaced by the 10 cm against the spring constant of $1000N{m^{ - 1}}$. For this, we will use the relation between the angular frequency and vertical displacement of the block along with the spring constant.

Complete step by step answer:
Mass of metal block, $m = 10kg$
Spring constant of the spring, $k = 1000N{m^{ - 1}}$
We know the maximum acceleration in a spring is given as $a = - {\omega ^2}y - - (i)$ where ‘y’ is the maximum displacement of the metal block from the mean position and $\omega$is the angular frequency.
According to the question, the maximum displacement of the block has been given as$y = 10cm = 0.1m$.
The square root of the ratio of the spring constant and the mass of the block results in the angular frequency of the block. Mathematically, $\omega = \sqrt {\dfrac{k}{m}} - - (ii)$
Substituting the value of the spring constant as 1000 newton per meter and the mass of the metal block to be 1 kilogram in the equation (ii) to determine the value of the angular frequency.

$$\Rightarrow\omega = \sqrt {\dfrac{k}{m}} \\\ \Rightarrow\omega = \sqrt {\dfrac{{1000}}{{10}}} \\\ \Rightarrow\omega = \sqrt {100} \\\ \Rightarrow\omega = 10{\text{ rad/s}} \\\$$

Hence, the value of the angular frequency is 10 radians per second.
Now, we substitute the value of angular frequency and the linear displacement of the spring in equation (i) to find the maximum acceleration.

$$\Rightarrow a = - {\omega ^2}y \\\ \Rightarrow a = - {\left( {10} \right)^2} \times 0.1 \\\ \Rightarrow a = - 100 \times 0.1 \\\ \therefore a =- 10m{s^{ - 1}} \\\$$

Hence, the maximum acceleration of the block is 10 meters per second.
Hence,option A is the correct answer.

Note: The fixed point (or position) of the block during its transition about which the block will move in to and fro motion is known as the equilibrium point. Here, it is interesting to note that the spring gets elongated due to the weight of the metal block.