সরল ছন্দিত স্পন্দনে যান্ত্রিক শক্তি

Suppose that we have an object attached to a spring in simple harmonic oscillation. We would like to know its kinetic and potential energy at any given point.

Define the following:

$m=$ mass of the object

$k=$ spring constant

$x=$ displacement from relaxed length

$X=$ maximum displacement

Suppose that an object is at position $x$. Let us calculate the work that the spring does in moving the mass to the relaxed position.

$$W=\int_{x}^{0}\vec{F}\cdot\vec{dx}$$ $$W=\int_{x}^{0}F\cdot dx$$

Now, we may simply use Hooke's Law: $$W=\int_{0}^{x}kx\cdot dx$$ $$W=\frac{1}{2}kx^2$$ This work done by the spring force is clearly the negative of the change in potential energy. Thus, the potential energy at a displacement $x$ is $U(x)=\frac{1}{2}kx^2$.

Now, let us take into account the maximum displacement, $X$. At this point, the kinetic energy is clearly $0$.

Now, we may calculate the kinetic energy at a point $x$ using the conservation of energy: $$\Delta K+\Delta U=0$$ $$\Delta K+\left(\frac{1}{2}kx^2-\frac{1}{2}kX^2\right)=0$$ $$K(x)-0=\frac{1}{2}kX^2-\frac{1}{2}kx^2$$ $$K(x)=\frac{k}{2}(X^2-x^2)$$ This is the kinetic energy at any given moment $x$.

If you are given parameters such as time and frequency, then you may use the function of position from time: $$x=X\cos(\omega t+\phi)$$ where $\omega=\sqrt{k/m}$ and $\phi$ is the phase angle. Of course, sine would also work.