Question 170: Oscillations

Question

At a given point of time the value of displacement of a simple harmonic oscillator is given as $$\mathrm{y}=\mathrm{A} \cos \left(30^{\circ}ight)$$. 
If amplitude is $$40 \mathrm{~cm}$$ and kinetic energy at that time is $$200 \mathrm{~J}$$, the value of force constant is $$1.0 	imes 10^{x} ~\mathrm{Nm}^{-1}$$. The value of $$x$$ is ____________.

Accepted Answer

Given the general equation for displacement in a simple harmonic oscillator:
$$x = A \sin(\omega t + \phi)$$
At the given time, we have:
$$\omega t + \phi = 30^\circ$$
Given the amplitude $$A = 40 \,	ext{cm}$$ and the displacement $$x = 40 	imes \frac{\sqrt{3}}{2} \,	ext{cm} = 20\sqrt{3} \,	ext{cm}$$, we can write the kinetic energy, $$KE$$, as:
$$KE = \frac{1}{2}k(A^2 - x^2) = 200$$
Now, substitute the values for $$A$$ and $$x$$:
$$200 = \frac{1}{2}k\left(\frac{1600 - 1200}{100 	imes 100}ight)$$
Simplify the equation:
$$400 	imes 100 	imes 100 = k 	imes 400$$
Solve for the force constant, $$k$$:
$$k = 10^4 \,	ext{Nm}^{-1}$$
Given that the force constant is expressed as $$k = 1.0 	imes 10^x \,	ext{Nm}^{-1}$$, comparing the values, we get:
$$1.0 	imes 10^x = 10^4$$
Thus, the value of $$x$$ is $$4$$.

Oscillations

Integer / Numeric Answer Type