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 3.37 Hz and an amplitude of 1.37 m. The particle oscillates between

x = 1.37 m

and

x = 1.37 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 1.48 s of this motion?
m

Explanation / Answer

X(t) = A*sin(wt)

A = 1.37

w = 2*pi*f = 2*pi*3.37 = 21.2 /s

X(t) = 1.37*sin(21.2t)

=============================

(b)

v = dx/dt = A*w*cos(wt)

Vmax = A*w

Vmax = A*w = 1.37*21.2 = 29 m/s

===================

acceleration a = dv/dt = -w^2*A*sin(wt)

a = -w^2*x

amax = -w^2*A

maximum acceleration amax = -21.2^2*1.37 = -615.7 m/s^2

=======================

time period T = 1/f = 1/3.37 = 0.297 s

distance covered in one time period = 4A

distance covered in t time = (4A/T)*t = (4*1.38*1.48/0.297 = 27.5 m <<<----------answer

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