A point mass initially located at x = 0 undergoes simple harmonic motion, initia

Question

A point mass initially located at x = 0 undergoes simple harmonic motion, initially the mass is moving in the positive x direction, with a frequency of 3.12 Hz and an amplitude of 1.13 m. The particle oscillates between

x = 1.13 m

and

x = 1.13 m.

(a) What is the equation describing the point mass's position as a function of time? (Do not include units in your answer.)

x(t) =

  

(b) What is the maximum speed of the particle?
m/s

(c) What is the maximum acceleration of the particle?
m/s2

(d) What is the total distance covered by the particle in the first 0.962 s of this motion?
m

Explanation / Answer

a)

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

where A = 1.13 m

f = 3.12 Hz

So, x(t) = 1.13*sin(2*pi*3.12*t)

x(t) = 1.13*sin(19.6*t)

b)

speed, v = dx/dt = 1.13*19.6*cos(19.6*t)

= 22.1*cos(19.6*t) m/s

So, maximum speed = 22.1 m/s <------answer

c)

Maximum acceleration = 22.1*19.6 = 433.2 m/s2

d)

time period, T = 1/f = 1/3.12 = 0.321 s

So, number of oscillations in 0.962s = 0.962/0.321 = 3

So, total distance covered = 4*A*3 = 4*1.13*3 = 13.6 m

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