In an engine, a piston oscillates with simple harmonic motion so that its positi

Question

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, 5) x = 8.00 cos 5t + where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At tO, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston Enter a number 2 erthe acceleration a(t) of an object if you know the velocity as a function of time, v(t)? cm/s (d) Find the period and amplitude of the motion. period amplitude cm

Explanation / Answer

given x = 8*cos(5t + pi/6)

comparing with the standard SHM x = A*cos(wt+phi)

amplitude A = 8 cm

angular frequency w = 5 s-1

phase constant phi = pi/6 rad

(a)

x = 8*cos((5*0 + pi/6)

x = 8*cos(pi/6)

x = 6.93 cm <<<------------ANSWER

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(b)

velocity v = dx/dt = -8*sin(5t + pi/6) *5

v = -5*8*sin( 5t + pi/6) = 40*sin(5t + pi/6)

at t= 0

v = -40*sin(5*0 + pi/6)

v = -20 cm/s <<<<---------ANSWER

(c)

acceleration a = dv/dt = -5*8*cos(5t + pi/6)*5

a = -25*8*cos(5t + pi/6)

a = -25*x

at t = 0 ,   x = 6.93

acceleration a = -25*6.93 = -173.25 cm/s^2

(d)

period T = 2pi/w = 2*pi/5 = 1.26 s

amplitude A = 8 cm

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