At a given point of time the value of displacement of a simple harmonic oscillator is given as  y  =  A cos(30∘). If amplitude is  40​cm and kinetic energy at that time is  200​J, the value of force constant is  1.0 × 10x​N​m−1. The value of  x is ______.

The problem involves a simple harmonic oscillator, which is a system that, when displaced from its equilibrium position, experiences a restoring force proportional to the displacement. The displacement of the oscillator is given by the equation y = A cos(30°), where A is the amplitude of the oscillation.

Given that the amplitude A is 40 cm, we can substitute this into the equation to find the displacement y at the given point in time:

y = 40 cos(30°)

The kinetic energy (K.E.) of the oscillator at this point in time is given as 200 J. The total energy of a simple harmonic oscillator is the sum of its kinetic and potential energies. At maximum displacement (i.e., the amplitude), the kinetic energy is zero and the potential energy is at a maximum. Therefore, the total energy is also the maximum potential energy, which is given by the equation:

P.E. = 1/2 k A^2

where k is the force constant. Since the total energy is equal to the kinetic energy at the given point in time, we can set the above equation equal to the given kinetic energy and solve for k:

200 J = 1/2 k (40 cm)^2

Solving this equation for k gives the force constant in N/m. The problem asks for the value of x in the expression 1.0 x 10^x N/m. To find x, take the base-10 logarithm of the force constant k.