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 2.00 cm, and the frequency is 1.50 Hz. (a) Find an expression for the position of the particle as a function of time. Determine (b) the maximum speed of the particle and (c) the earliest time (t > 0) at which the particle has this speed. Find (d) the maximum positive acceleration of the particle and (e) the earliest time (t > 0) at which the particle has this acceleration. (f) Find the total distance traveled by the particle between t = 0 and t = 1.00 s.

Explanation / Answer

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

here at t = 0 , x = 0 ;

so x = ( 0.02 ) cos ( t + ) ;

= 2f = 2 ( 1.5) = 3 ;

0 = ( 0.02 ) cos ( 2* 0 + ) ;

or = 90 degrees or /2 radians

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

(b)maximum speeed is A = 0.02 * ( 3 ) = 6 cm/s

(c) v = - A sin ( t + /2  )

or - 6 = - ( 0.02 ) ( 3 ) sin ( 3 t + /2  ) ;

maximum velocity occurs at centre , i e at T /2 or 1/ ( 2 f ) = 0.333s ;

(d) maximum positive acceleration = A ^2 = 2 * ( 3 ) ^2 = 182cm / s^2 ;

(e) earliest time , at t = 3T / 4 , or t = 3/( 4 * 1.5 ) = 0.500 s

( f ) time period = T = 1 / f = 1/ 1.5 = 0.666 s ,

number of time periods t=1s = 1.5T

That means it travels 0 -> A -> - A -> A ->0, for a

d = A + 2A + 2A + A = 6A = 12:0 cm.

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