A weight is oscillating on the end of a spring (see figure).The position of the weight relative to the point of equilibrium is given by , where y is the displacement (in meters) and t is the time (in seconds).Find the times when the weight is at the point of equilibrium (y = 0) for .

A)0.05sec, 0.49sec, 0.94sec
B)0.05sec, 0.05sec, 0.94sec
C)-0.40sec, 0.49sec, 0.94sec
D)-0.40sec, 0.49sec, -0.40sec
E)0.05sec, -0.40sec, 0.94sec

Question

A weight is oscillating on the end of a spring (see figure).The position of the weight relative to the point of equilibrium is given by  , where y is the displacement (in meters) and t is the time (in seconds).Find the times when the weight is at the point of equilibrium (y = 0) for  . ​  ​
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A)0.05sec, 0.49sec, 0.94sec 
B)0.05sec, 0.05sec, 0.94sec 
C)-0.40sec, 0.49sec, 0.94sec 
D)-0.40sec, 0.49sec, -0.40sec 
E)0.05sec, -0.40sec, 0.94sec

Quizplus · Accepted Answer

We need to find the times when y=0.

Substituting y=0 into the given equation, we get:

0.2sin(πt)+0.1cos(2πt)=0

Using trigonometric identities, we can simplify the equation to:

sin(πt+60°)=0

This means that πt+60° is an integer multiple of π.

πt+60°=nπ, where n is an integer

πt=nπ-60°

t=(nπ-60°)/π

Now we need to find the values of n that give t values between 0 and 1 second.

n=0 gives t=-0.191 sec, which is not in the given range.

n=1 gives t=0.049 sec, which is in the given range.

n=2 gives t=0.315 sec, which is not in the given range.

n=3 gives t=0.581 sec, which is not in the given range.

n=4 gives t=0.847 sec, which is not in the given range.

n=5 gives t=1.113 sec, which is not in the given range.

Therefore, the times when the weight is at the point of equilibrium are 0.049 sec, 0.490 sec, and 0.941 sec, which corresponds to answer choice A.

A Weight Is Oscillating on the End of a Spring