Find the amplitude of the oscillation of the particle.

Question

The displacement of a particle varies according to $x=3(\cos t +\sin t)$. Then find the amplitude of the oscillation of the particle.

Can someone kindly explain the concept of amplitude and oscillation and how to solve it?

Any hints for solving the problem would be helpful.

Answer 1

The amplitude of a particle is the largest distance it moves from the equilibrium point when moving periodically. Oscillation is just a word used to define movement which follows a regular pattern.

To find the amplitude of the oscillation of $x$ we need to find the maximum value of $3\cos{t}+3\sin{t}$ when $t$ varies. $$3\cos{t}+3\sin{t}=3\sqrt{2}\sin{(t+\frac{\pi}{4})}$$ So, the maximum value of the displacement - the amplitude of the particle - is $3\sqrt{2}$.

Answer 2

The amplitude is $\boxed{3\sqrt{2}}$

In general, $\alpha \cos(x) + \beta \sin(x)$ has amplitude given by $\sqrt{\alpha^{2} + \beta^{2}}$. Here, you have $\alpha = \beta = 3$, which gives $\sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$.

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The period is $\boxed{2\pi}$ since this is just the sum of two standard trig functions. The argument of both are $x$.

Answer 3

The displacement of a particle given by $x(t)=3(\cos t+ \sin t)$ can be writren as a sinusoid. $$x(t) = 3\sqrt 2 (\frac{\cos t}{\sqrt 2} + \frac{\sin t}{\sqrt 2})$$

$$x(t)=3\sqrt 2 \sin( t+\pi/4)$$

Since the displacement of the particle can be written as a sinusoid , the motion is oscillatory. The amplitude is simply the maximum displacement of the particle which is nothing but $$A=3\sqrt 2$$

This can also be visualized as a rotating vector whose projection on the y axis will be oscillatory. The magnitude of such a rotating vector is the amplitude of the oscillation.

Just differentiate the displacement twice to check whether $$a(t)\propto -x(t)$$

Then the particle motion is oscillatory. Hope this helps...