An oscillating mass spring system has a mechanical energy $1\,J$, when it has an amplitude $0.1\,m$ and maximum speed of $1\,m{s^{ - 1}}$. Then the force constant of the spring is
(A) $100\,N{m^{ - 1}}$
(B) $200\,N{m^{ - 1}}$
(C) $300\,N{m^{ - 1}}$
(D) $50\,N{m^{ - 1}}$

Hint:The force constant can be determined by using the Simple harmonic motion formula. This formula gives the relation between the force constant and the amplitude. By using the given information, and by using the simple harmonic motion formula of spring, the force constant can be determined.

Formulae Used:
The simple harmonic motion of spring,
$E = \dfrac{1}{2}K{A^2}$
Where, $E$ is the energy of the spring, $K$ is the force constant of the spring and $A$ is the amplitude of the spring.

Complete step-by-step solution:
Given that,
The mechanical energy of the spring is, $E = 1\,J$
The amplitude of the spring, $A = 0.1\,m$
The maximum speed of the spring is $1\,m{s^{ - 1}}$
Now, the simple harmonic motion of spring,
$E = \dfrac{1}{2}K{A^2}\,....................\left( 1 \right)$
By substituting the energy of the spring amplitude of the spring in the above equation (1), then the above equation is written as
$1 = \dfrac{1}{2}K{\left( {0.1} \right)^2}$
On squaring the term inside the bracket, then the above equation is written as,
$1 = \dfrac{1}{2}K \times 0.01$
By keeping the term $K$ in one side and the other terms in other side, then the above equation is written as,
$K = \dfrac{2}{{0.01}}$
On dividing the above equation, then the above equation is written as,
$K = 200\,N{m^{ - 1}}$
Thus, the above equation shows the force constant of the spring is $200\,N{m^{ - 1}}$.
Hence, the option (B) is the correct answer.

Note:- The unit of force constant speed is given by the calculation itself. The unit of energy is newton meter and the unit of amplitude is meter by squaring the amplitude the unit will become meter square. Then by dividing the newton metre by meter square, then the unit will become newton per metre.