Question

Establish the differential equation for a system executing simple harmonic motion
(SHM). Show that, for SHM, the velocity and acceleration of the oscillating object is
proportional to 0
w and 2
0
w , respectively, where 0
w is the natural angular frequency of
the object.

Answer 1

Answer on Question #41966, Physics, Mechanics | Kinematics | Dynamics

Establish the differential equation for a system executing simple harmonic motion (SHM). Show that, for SHM, the velocity and acceleration of the oscillating object is proportional to $\omega_0$ and $\omega^2_0$ , respectively, where $\omega_0$ is the natural angular frequency of the object.

Solution:

Simple harmonic motion is typified by the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law.

Now since $F = -kx$ is the restoring force and from Newton's law of motion force is given as

$$F = m a,$$

where $m$ is the mass of the particle moving with acceleration $a$ . Thus acceleration of the particle is

$$a = \frac {F}{m} = \frac {- k x}{m}$$

but we know that acceleration $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

Thus,

$$\frac {d ^ {2} x}{d t ^ {2}} = \frac {- k x}{m}$$

This differential equation is known as the simple harmonic equation.

The solution is

$$x = \cos (\omega_ {0} t + \phi)$$

where $A, \omega_0$ and $\phi$ are all constants.

We know that velocity of a particle is given by

$$v = \frac {d x}{d t}$$

Now differentiating the displacement of particle $x$ with respect to $t$

$$v = \frac {d x}{d t} = \omega_ {0} (- \sin (\omega_ {0} t + \phi))$$

From trigonometry we know that

$$\sin^ {2} x + \cos^ {2} x = 1$$

Thus,

$${}^2 \sin^2(\omega_0 t + \phi) = {}^2 - {}^2 \cos^2(\omega_0 t + \phi) = {}^2 - x^2$$

Or

$$\sin(\omega_0 t + \phi) = \sqrt{1 - \frac{x^2}{2}}$$

putting this in for velocity we get,

$$v = -A \omega_0 \sqrt{1 - \frac{x^2}{2}}$$

so it is proportional to $\omega_0$.

Again we know that acceleration of a particle is given by

$$a = \frac{dv}{dt} = \frac{d}{dt} \left(-A \omega_0 \sin(\omega_0 t + \phi)\right) = -A \omega_0^2 \cos(\omega_0 t + \phi) = -\omega_0^2 x$$

so it is proportional to $\omega^2$.

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