A particle moving along the x axis in simple harmonic motion starts from its equ

Question

A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 1.50 cm, and the frequency is 1.70 Hz.

(a) Find an expression for the position of the particle as a function of time. (Use the following as necessary: t. Assume that x is in centimeters and t is in seconds. Do not include units in your answer.)
x =_______

(b) Determine the maximum speed of the particle.
_______cm/s

(c) Determine the earliest time (t > 0) at which the particle has this speed
_______s

(d) Find the maximum positive acceleration of the particle.
_______cm/s2

(e) Find the earliest time (t > 0) at which the particle has this acceleration.
_______s

(f) Determine the total distance traveled between t = 0 and t = 0.88 s.
_______cm

Explanation / Answer

a)

the general equation of motion of SHM :

x = A*sin(2*pi*f*t) =1.5*sin(2*pi*1.7*t)

So, x = 1.5*sin(10.7*t)

b)

maximum speed = A*2*pi*f

= 1.5*2*pi*1.7

= 16.02 cm/s

c)

For earliest time,t = T/2

where T = time period = 1/f

So, t = 1/2f = 1/(2*1.7) = 0.294 s

d)

maximum positive acceleration = A*(2*pi*f)^2

= 1.5*(2*pi*1.7)^2

= 171.1 cm/s2

e)

earliest time, t = 3T/4 = 3/4f

= 3/(4*1.7)

= 0.441 s

f)

total distance traveled = 6A = 6*1.5 = 9 cm

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