A particle initially located at the origin undergoes simple harmonic motion, mov

Question

A particle initially located at the origin undergoes simple harmonic motion, moving first in the positive z direction, with a frequency of 3.47 Hz and an amplitude of 1.35 m.The particle oscillates between z = 1.35 m and z = 1.35 m.

(a) What is the equation describing the particle's position as a function of time? (Use the following as necessary: t. Do not include units in your answer. Assume t is in seconds.)

z(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 1.15 s of this motion?
_____m

Explanation / Answer

here,

frequency , f = 3.47 Hz

angular frequency , w = 2*pi*f

w = 6.28 * 3.47

w = 21.79 rad/s

amplitude of oscillation , A= 1.35 m

(a)

the equation describing the particle's position as a function of time is

z(t) = A * sin(w*t)

at t = 0 ,the particle is at origin

t is in seconds

(b)

the maximum speed of the particle , vmax= A*w

vmax = 29.42 m/s

(c)

the maximum acceleration of the particle , amax = A*w^2

amax = 640.99 m/s^2

(d)

at t = 1.15 s

z(1.15) = 1.35 *sin(21.79*1.15 )

= - 0.1 m

the total distance covered by the particle in the first 1.15 s of this motion x = 4*A - 0.1 m

x = 5.3 m

the total distance covered by the particle in the first 1.15 s of this motion is 5.3 m

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