A weight attached to a spring is at its lowest point, 9 inches below equilibrium, at time $t=0$ seconds. When the weight it released, it oscillates and returns to its original position at $t=3$ seconds. Which of the following equations models the distance, $d$, of the weight from its equilibrium after $t$ seconds?
$d=-9 \cos \left(\frac{\pi}{3} t\right)$
$d=-9 \cos \left(\frac{2 \pi}{3} t\right)$
$d=-3 \cos \left(\frac{\pi}{9} t\right)$
$d=-3 \cos \left(\frac{2 \pi}{9} t\right)$
Answer (1)
The problem involves finding the equation of oscillatory motion of a weight attached to a spring.
The general form of the equation is d = A cos ( Bt ) , where A is the amplitude and B is related to the period.
Given that the weight is at its lowest point 9 inches below equilibrium at t = 0 , the amplitude A = − 9 .
The period is 6 seconds, so B = 3 π , and the equation is d = − 9 cos ( 3 π t ) .
Explanation
Problem Analysis The problem describes a weight oscillating on a spring, starting at its lowest point. We need to find the equation that models the distance d of the weight from its equilibrium after t seconds.
General Equation Since the weight starts at its lowest point, we can model the distance using a cosine function with a negative amplitude. The general form of the equation is: d = A cos ( Bt ) where:
A is the amplitude (the maximum displacement from equilibrium).
B is related to the period of oscillation.
Determining Amplitude The weight is at its lowest point, 9 inches below equilibrium, at t = 0 . This means the amplitude A = − 9 inches. So the equation becomes: d = − 9 cos ( Bt )
Calculating B The weight returns to its original position at t = 3 seconds. This means that half of the period is 3 seconds, so the full period T is 2 × 3 = 6 seconds. The relationship between the period T and B is: T = B 2 π Substituting T = 6 :
6 = B 2 π Solving for B :
B = 6 2 π = 3 π
Final Equation Now we have A = − 9 and B = 3 π . Substituting these values into the general equation, we get: d = − 9 cos ( 3 π t )
Matching the Equation Comparing this equation with the given options, we find that it matches the first option: d = − 9 cos ( 3 π t )
Examples
Understanding oscillatory motion is crucial in many fields, such as physics and engineering. For example, when designing suspension systems for cars, engineers use equations similar to the one we derived to model the behavior of the springs and dampers. By accurately modeling the oscillations, they can optimize the system to provide a smooth and comfortable ride. Similarly, in electrical engineering, oscillatory motion is used to describe alternating current (AC) circuits, where the voltage and current oscillate sinusoidally. The equation we found can be used to model the voltage or current in an AC circuit, allowing engineers to analyze and design these circuits effectively.
