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 3.00 cm, and the frequency is 1.00 Hz.
(a) Find an expression for the position of the particle as a function of time. (Use the following as necessary: t, and p.)
x =

(b) Determine the maximum speed of the particle.
answer in cm/s

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

(d) Find the maximum positive acceleration of the particle.
answer in cm/s^2

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

(f) Determine the total distance traveled between t = 0 and t = 1.50 s.
answer in cm

Explanation / Answer

(a) x = A cos ( t + ) ;

here at t = 0 , x = 0 ;

so x = ( 0.03 ) cos ( t + ) ;

= 2f = 2 ( 1 ) = 2 ;

0 = ( 0.03 ) cos ( 2* 0 + ) ;

or = 90 degrees or /2 radians

x = ( 0.03 ) cos ( t + /2 ) ;

(b)maximum speeed is A = 0.03 * ( 2 ) = 0.1884955 m /s = 18.8495 cm / s ;

(c) v = - A sin ( t + /2  )

or - 0.1884955 = - ( 0.03 ) ( 2 ) sin ( 2 t + /2  ) ;

maximum velocity occurs at centre , i e at T /2 or 1/ ( 2 f ) = 0.5 * 1 = 0.5s ;

(d) maximum positive acceleration = A ^2 = 3 * ( 2 ) ^2 = 118.4 cm / s^2 ;

(e) earliest time , at t = T / 4 , or t = 1 /( 4 * 1 ) = 0.25 s

( f ) time period = T = 1 / f = 1/ 1 = 1 s ,

number of time periods = 1.5 / 1 = 1.5 time periods ,

distance in one time period = 3 *4 =12 cm ;

distance in 1.5 s = 12 * 1.5 = 18 cm

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