Question

Physics For Scientists And Engineers (3rd Edition)

Chapter 14, Problem 10E

I am lost on the begining of step 2.

How exactly did phi get isolated?

An air track glider attached to a spring oscillates with a period of 1.5 seconds. At t=0 the glider is 5.0 cm to the left of equilibrium and moving to the right at 36.3 m/s.

What is the phase constant?

What is the phase at t= 0, .5, 1.0, 1.5

Thanks

Answer 1

Solution:

Let's assume x(t) = Asin(?t + ?)
Then v(t) = A?cos(?t + ?)

x(0) = 5 cm = Asin(?*0 + ?)
v(0) = 36.3 cm/s = A?cos(?*0 + ?)

v(0) / x(0) = 36.3cm/s / 5cm = 7.26 rad/s = A?cos? / Asin? = ? / tan?
But ? = 2?/T = 2? / 1.5s ? 4.2 rad/s
tan? = 4.2rad/s / 7.26rad/s = 0.577
? = 30º = ?/6

b) Not sure what you mean by phase. If you mean "the expression inside the trig function", then
phase(0) = ?/6
phase (0.5s) = ?t + ?/6 = 2?(0.5s)/1.5s + ?/6 = 2?/3 + ?/6 = 5?/6
phase (1.0s) = 2?(1/1.5) + ?/6 = 8?/6 + ?/6 = 3?/2
phase (1.5s) = 2? + ?/6 = 13?/6 = ?/6

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