Question

An object in simple harmonic motion has amplitude 4.0{\rm cm} and frequency 3.6{\rm Hz} , and at t = 0\;{\rm s} it passes through the equilibrium point moving to the right. Write the function x(t) that describes the object's position.

Write the function x(t) that describes the object's position. x(t) =( 4.0{\rm cm} )\cos((7.2\pi \;{\rm rad/s}) t -\pi /2 \;{\rm rad}) x(t) =(2.0{\rm cm} )\cos((7.2\pi \; {\rm rad/s}) t -\pi /2 \;{\rm rad}) x(t) = (4.0{\rm cm} )\cos((3.6 \;{\rm rad/s} )t) x(t) =(2.0{\rm cm} )\cos((7.2\pi \; {\rm rad/s}) t)

Answer 1

Answer:

(A) x(t) = (4.0~{\rm cm})\cos(7.2\pi - \pi/2)

Explanation:

The general equation of motion in simple harmonic motion is as follows:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and Ф is the phase angle which is to be determined by initial conditions.

The frequency is given in the question, so the angular frequency is

$\omega = 2\pi f = 2\pi (3.6) = 7.2\pi$

$A = 4\times 10^{-2}m$

The initial condition is at t = 0, the object passes through the equilibrium to the right.

$x(t=0) = A\cos(\phi) = 0$

In order the cosine term to be zero it should satisfy the following condition:

$\phi = \frac{n\pi}{2}$

where n is an integer.

The object is moving right (+x-direction), so its velocity is positive at t = 0.

The velocity function is the derivative of the position function:

$v(t) = -\omega A\sin(\omega t + \phi)$

$v(t=0) = -\omega A\sin(\phi) > 0$

Then, sin(Ф) should be negative, therefore Ф should be negative.

Finally, the equation of motion can be written:

$x(t) = A\cos(\omega t - \pi/2)\\x(t) = (4.0~{\rm cm})\cos(7.2\pi - \pi/2)$

Answer 2

The function x(t) that describes the object's position is equal to $x(t)=4cos (7.2\pi-\frac{\pi}{2} )$

Given the following data:

Amplitude = 4.0 cm.

Frequency = 3.6 Hz.

Time = 0 m/s.

How to calculate the position of an object in simple harmonic motion.

Mathematically, the standard equation for the position of an object in simple harmonic motion is given by:

$x(t)=Acos(\omega t+\phi)$

Where:

• A is the amplitude.

• $\omega$ is the angular frequency.

• $\phi$ is the phase angle.

But, angular frequency is given by:

$\omega = 2\pi f\\\\\omega = 2\pi \times 3.6\\\\\omega = 7.2\pi$

For the phase angle, we have:

Note: At t = 0, the object is moving towards the right.

$V(t)=-\omega Asin(\omega t +\phi)\\\\V(0)=-\omega Asin \phi > 0\\\\\phi=\frac{-\pi}{2}$

Therefore, the function x(t) that describes the object's position is equal to $x(t)=4cos (7.2\pi-\frac{\pi}{2} )$.

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