Question

An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the amplitude and (b) the period of this motion.

Answer 1

(1) The position around equilibrium of an object in simple harmonic motion is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the motion
$\omega$ is the angular frequency.

The velocity is the derivative of the position:
$v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)$
where 
$v_0 = \omega A$ is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
$a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)$
where
$a_0=\omega^2 A$ is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
$v_0 = \omega A = 4.3 m/s$
$a_0 = \omega^2 A = 0.65 m/s^2$
From the first equation, we get
$A= \frac{4.3 }{\omega}$ (1)
and if we substitute this into the second equation, we find the angular fequency
$\omega=0.15 rad/s$
while the amplitude is  (using (1)):
$A=28.7 m$


(b) We found in the previous step that the angular frequency of the motion is
$\omega=0.15 rad/s$
But the angular frequency is related to the period by
$T= \frac{2 \pi}{\omega}$
and so, the period is
$T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s$

Answer 2

The amplitude will be equal to A=28.7 m and the time period of the object will be equal to T=41.9 s

What is Simple harmonic motion?

Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.

1) The position around equilibrium of an object in simple harmonic motion is described by

$x(t)=Acos(wt)$

where

A is the amplitude of the motion

w is the angular frequency.

The velocity is the derivative of the position:

$v(t)=-wAsinwt=-v_osinwt$

where

$v_o=wA$ is the maximum velocity of the object.

The acceleration is the derivative of the velocity:

$a(t)=-w^2Acoswt=-a_ocoswt$

where

$a_o=w^2A$is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:

$v_o=wA=4.3 m/s$

$a_o=w^2A=0.65\ m/s^2$

From the first equation, we get

$A=\dfrac{4.3}{w}$     (1)

and if we substitute this into the second equation, we find the angular fequency

$w=0.15 rad/s$

while the amplitude is  (using (1)):

A=27.4 m

(b) We found in the previous step that the angular frequency of the motion is

$w=0.15 rad/s$

But the angular frequency is related to the period by

$T=\dfrac{2\pi}{w}$

and so, the period is

$T=\dfrac{2\pi}{0.15}=41.9\ sec$

Thus the amplitude will be equal to A=28.7 m and the time period of the object will be equal to T=41.9 s

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