Question

An air-track glider attached to a spring oscillates with a period of 1.50 s . At t=0s the glider is 5.30 cm left of the equilibrium position and moving to the right at 38.5 cm/s .

A.) What is the phase constant?

B.) What is the phase constant at t=1.5s?

Answer 1

A.) Since Equation of oscillation is given by,

x(t) = A*sin(w*t + $\phi$) eq(1)

here, x(t) = Position

A = Amplitude

t = time

w = 2*pi/T

T = time period

$\phi$ = phase constant

at, t = 0

x(0) = -5.30

w = 2*pi/1.50 = 4*pi/3

So, -5.30 = A*sin(w*0 + $\phi$)

-5.30 = A*sin$\phi$

now differentiate eq(1) with respect to 't',

dx(t)dt = v(t) = A*w*cos(w*t + $\phi$)

at, t = 0

v(0) = +38.5 = A*w*cos(w*0 + $\phi$)

38.5 = A*(4*pi/3)*cos$\phi$

By dividing bold equations,

-5.30/38.5 = [1/(4*pi/3)]*tan$\phi$

tan$\phi$ = (4*pi/3)*(-5.30/38.5)

$\phi$ = arctan[(4*pi/3)*(-5.30/38.5)]

$\phi$ = -29.97 deg

$\phi$= -pi/6 rad

B.)

Phase constant at time 't' is given by,

$\phi$(t) = w*t + $\phi$

given, t = 1.5 s

So, $\phi$(t) = (4*pi/3)*1.5 + (-pi/6) = 2*pi - pi/6

$\phi$(t) = 11*pi/6

Please upvote.