A mass is attached to a hanging spring and undergoes simple harmonic motion about an equilibrium location at y = 0. The displacement of the mass (in cm) is given by y=10cos(πt/6) where t is the number of seconds since the mass was initially pushed or pulled from equliibrium. Assume positive displacement means the mass is above equilibrium.
List the first TWO times the mass passes through the equilibrium location, separated by commas.
To find the times when the mass passes through the equilibrium location, we need to find the values of t for which y = 0.
Given the equation for displacement y = 10cos(πt/6), we can set it equal to zero and solve for t:
0 = 10cos(πt/6)
Now, divide both sides by 10:
0 = cos(πt/6)
Next, we solve for t by taking the inverse cosine (or arccos) of both sides:
cos^(-1)(0) = πt/6
The inverse cosine of zero is π/2:
π/2 = πt/6
To solve for t, we multiply both sides by 6/π:
t = 6/π * (π/2)
Simplifying the expression, we have:
t = 3
So, the first time the mass passes through the equilibrium location is at t = 3 seconds.
To find the second time, we need to find the next value of t where y = 0. Since the sine function is periodic with a period of 2π, the next time the mass passes through the equilibrium will be another π/2 radians later. Therefore, the second time will be:
t = 3 + (π/2) = 3 + 1.57 = 4.57 seconds (rounded to two decimal places)
So, the first two times the mass passes through the equilibrium location are 3 seconds and 4.57 seconds.