An air-track glider is attached to a spring. The glider is pulled to the right a

Question

An air-track glider is attached to a spring. The glider is pulled to the right and released from rest att=0s. It then oscillates with a period of 1.5 s and a maximum speed of 42 cm/s . Part A What is the amplitude of the oscillation? I did this already using vmax = wA. w = 2 pief and f is 1/T so A is 10cm.

Part B What is the glider's position at t= 30 s ?

Could you show each step and tell me how to enter this in my calculator?

I looked at x(t) = Acos(wt). However, here t is at 0 so would need to use sine I think.

so 10sin (wt) and w = 2Pief with f =1/T so w is 2 pie/T or 4.188 sec.

but I can't get an answer that looks rights with 10 sin (4.188 x 30sec). or 10 sin (125.6637). Could you show steps and how I'd enter it into my calculator

Explanation / Answer

given that

T = 1.5 sec

vmax = 42 cm/s = 0.42 m/s

t = 30 sec

part (A)

x = a cos (w*t)

a=amplitude

w = 2*pi / Period = angular speed

check equ with given conditions

v = speed = dx/dt = d(acos(w*t))/dt = -a w sin(wt)

t=0, v = 0 because sin (0) =0 because given released from rest

max speed v(max) will be at mean position (x=0)     so from above equation

0 = a cos (wt)

cos (wt) = 0

so sin (wt) =1 at v(max)

|vmax| = a*w       [by putting sin(w*t) = 1]

0.42 = (a *2*pi ) / T = a *2*pi/1.5                 (we know w = 2*pi/T)

a = (0.42*1.5) / (2*3.14) =  0.100 meter

amplitude = 10.0 cm

part (B)

x = a cos (w*t)

at given t = 30 s

x = 10 cos [2*pi *30/1.5]

x = 10 cos [2*3.14*30/1.5] = 10*-0.58

x = -5.8 cm

the negative sign show that the glider is 5.8 cm on the LEFT of the equilibrium point.

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