A weight is oscillating on the end of a spring. The position of the weight relative to the point of equilibrium is given by y=1/12(cos8t-3sin8t), 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 0<=t<=1.

Question

A weight is oscillating on the end of a spring. The position of the weight relative to the point of equilibrium is given by y=1/12(cos8t-3sin8t), 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 0<=t<=1.

Answer 1

when y=0

0=1/12 (cos8t-3sin8t)

or
cos8t=3sin8t
tan8t=1/3
8t=arctan 1/3
t= 1/2 arctan 1/3

check my thinking.

Answer 2

To find the times when the weight is at the point of equilibrium (y=0), we need to solve the equation y=0.

Let's rewrite the equation:
y = (1/12)(cos(8t) - 3sin(8t))

To find when y=0, we set this equation equal to zero and solve for t:
0 = (1/12)(cos(8t) - 3sin(8t))

First, let's multiply both sides by 12 to remove the fraction:
0 = cos(8t) - 3sin(8t)

Now, we can rewrite sin(8t) using the identity sin(8t) = cos(π/2 - 8t):
0 = cos(8t) - 3cos(π/2 - 8t)

Next, we can use the identity cos(A - B) = cos(A)cos(B) + sin(A)sin(B) to simplify the equation further:
0 = cos(8t) - 3cos(π/2)cos(8t) + 3sin(π/2)sin(8t)

Since cos(π/2) = 0 and sin(π/2) = 1, the equation becomes:
0 = cos(8t) - 3(0)cos(8t) + 3(1)sin(8t)
0 = cos(8t) + 3sin(8t)

Now, let's use the angle addition identity for cosine:
0 = cos(8t) + 3sin(8t)
0 = cos(8t)(1 + 3tan(8t))

For the equation to be true, either cos(8t) = 0 or 1 + 3tan(8t) = 0.

Let's first solve cos(8t) = 0:
cos(8t) = 0

To find the values of t for which cos(8t) = 0, we look for the values of 8t that correspond to the x-intercepts of the cosine function. The cosine function has an x-intercept at x = (2n + 1)π/2, where n is an integer.

So, we have 8t = (2n + 1)π/2.

Now, let's solve 1 + 3tan(8t) = 0:
1 + 3tan(8t) = 0

To find the values of t for which 1 + 3tan(8t) = 0, we solve for t using inverse trigonometric functions. Taking the arctangent (tan^(-1)) of both sides gives us:
tan^(-1)(-1/3) = 8t

Now, we solve for t by dividing both sides by 8:
t = tan^(-1)(-1/3) / 8

To find the times when the weight is at the point of equilibrium (y=0) for 0<=t<=1, we substitute the values of t we found into the original equation:
t = tan^(-1)(-1/3) / 8
t = (2n + 1)π/16

By substituting different values of n, we can find the specific times when the weight is at the point of equilibrium within the given time interval.