Question

In Problems 40 and 41, the distance d (in feet) that an object travels in time t (in seconds) is given.

(a) Describe the motion of the object.

(b) What is the maximum displacement from its rest position?

(c) What is the time required for one oscillation?

(d) What is the frequency?

$d=6\sin \left(2t\right)$

Short answer

(a) The motion is harmonic.

(b) Maximum distance covered from initial position is 6 feet.

(c) Time period to cover one oscillation is $\mathrm{\pi }$seconds.

(d) The frequency islocalid="1646895030365" $\frac{1}{\mathrm{\pi }}$

Step-by-step solution

Part (a) Step 1. Given Information

The equation of distance with respect to time is$d=6\sin \left(2t\right)$.

Part (a) Step 2. Compare the Equation with Equation of Harmonic Motion

• The equation of simple harmonic motion is $d=a\sin \left(\omega t\right)$.
• The given equation is same as this equation with $a=6,\omega =2$
• So, the motion of the object is simple harmonic motion.

Part (b) Step 1. Given Information

The equation of distance is$d=6\sin \left(2t\right)$.

Part (b) Step 2. Find Maximum Distance from starting position

• The maximum distance is$\left|a\right|$.
• Here, $a=6$.
• So, the maximum distance is 6 feet.

Part (c) Step 1. Given Information

The given equation of distance is$d=6\sin \left(2t\right)$.

Part (c) Step 2: Find the Time Period of 1 Oscillation

• The time period is given by $\frac{2\mathrm{\pi }}{\omega }$.
• Here, $\omega =2$.
• So, the time period of one oscillation is$\frac{2\mathrm{\pi }}{2}=\mathrm{\pi }$

Part (d) Step 1: Given Information

The equation of distance covered by the object is $d=6\sin \left(2t\right)$.

Part (d) Step 2: Find Frequency

• The frequency is given by the reciprocal of the time period of one oscillation.
• So, the frequency is $f=\frac{1}{\mathrm{\pi }}$oscillation per seconds.