Question

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x= 5.0 cos (2t + 3.1416/6 Where x is in centimeter and t is seconds. At t=0, Find a) the position of the particle b) its angular velocity and frequency c)its acceleration d)The period e)the amplitude of the motion

Answer 1

a) x = 4.33 m , b) w = 2 rad / s , f = 0.318 Hz , c) a = - 17.31 cm / s²,

d) T = 3.15 s, e) A = 5.0 cm

Step-by-step explanation:

In this exercise on simple harmonic motion we are given the expression for motion

x = 5 cos (2t + π / 6)

they ask us for t = 0

a) the position of the particle

x = 5 cos (π / 6)

x = 4.33 m

remember angles are in radians

b) The general form of the equation is

x = A cos (w t + Ф)

when comparing the two equations

w = 2 rad / s

angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 2 / 2pi

f = 0.318 Hz

c) the acceleration is defined by

a == d²x / dt²

a = - A w² cos (wt + Ф)

for t = 0 , we substitute

a = - 5,0 2² cos (π / 6)

a = - 17.31 cm / s²

d) El period is

T = 1/f

T= 1/0.318

T = 3.15 s

e) the amplitude

A = 5.0 cm