Period in simple harmonic motion [closed]

Question

I would like to know why period in simple harmonic motion does not depend on the amplitude of oscillation.

Answer 1

The amplitude increases with the speed. The period is proportional to the amplitude, but inverse proportional to the speed. So, because the amplitude is proportional to the speed, their influences on the period cancel out.

Answer 2

Simple harmonic motion obeys the differential equation $\ddot{x}+\omega^2 x=0$ for a constant $\omega>0$ with the units of frequency. Since this is a homogeneous linear equation, its solutions are closed under multiplication by constants. Doing this changes the amplitude, but not the frequency.

Answer 3

In the general case, there is a body connected with a spring, as in the picture:
Let's assume that there is no friction. The restoring force is the only force that is acting on the body and when we apply Newton's second law we get:

$ma = -kx$

Then we get:

$ma + kx = 0$

This equation can take the following form:

$m\frac{d^2x}{dt^2} + kx = 0$ $(a = \frac{d^2x}{dt^2})$

As acceleration is defined as the derivative of velocity, v, with respect to time t and velocity is defined as the derivative of position, x, with respect to time, acceleration can be thought of as the second derivative of x with respect to t. This is just a more complicated way to say that acceleration shows the rate of change of position over time. Calculous is required to answer your question.

$\frac{d^2x}{dt^2} + \frac{k}{m}x = 0$

Now we can substitute $\frac{m}{k} = \omega_o^2$ where is angular frequency. There is one more formula regarding angular frequency: $T = \frac{2\pi}{\omega_o}$ and finaly we get that $T = 2\pi\sqrt{\frac{k}{m}}$. You can see that the period doesn't depend on the amplitude beacuse it is not connected with it over the whole proof.

Best regards,
Sinestro