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 is2.50 cm, and the frequency is 1.30 Hz.

(a) Find an expression for the position of the particle as a function of time. (Use the following as necessary: t, and ?.)
x =  

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

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

(d) Find the maximum positive acceleration of the particle.
cm/s2

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

(f) Determine the total distance traveled between t = 0 and t = 1.15 s.
cm

the second Q

How much energy is required to move a 900 kg object from the Earth's surface to an altitude twice the Earth's radius?

Explanation / Answer

amplitude A =2.50 cm =0.025 m
freq f = 1.30 Hz
ang ferq w = 2pi*f = 6.28*1.30 = 8.164 rad/s
a)
x = 0.025 sin(8.164 t) m
--------------------------------
b)
maximum speed v = wA = 8.164*0.025 = 0.2041 m/s
--------------------------------
c)
at x =0
sin 8.164 t = 0
8.164 t= pi
t = 3.14/8.164 =0.384 sec
-------------------------
d)
max acceleration = w^2 *A = 8.164^2*0.025 = 1.6662724 m/s^2
-----------------------
e)
at t = 3T/4 = 3/4f = 3/4*1.30 = 0.5769 sec
-------------------------------
f)peroid T = 1/1.30 =0.769 sec
for T peroid the distance travelled is 4A
t = 1.15 sec
that is in the time of 1.5T
distance travelled is 6A = 6(0.025) = 0.15 m

--------------------------------------------
Energy E = U2-U1 = -GMm/(3R) -(-GMm/R) = -2GMm/3R = -2*6.67x10^-11*6*10^24*900/(3*6400x10^3)
E = 3.75 x 10^10 J

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