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?

Answer 1

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
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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

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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

Answer 2

its motion should be in the form :

x(t) = 2.1 sin (2pi f) t

=> x(t) = 2.1 sin 8.17 t

so its speed is :

v(t) = dx/dt = 8.17(2.1) cos 8.17t

=> v(t) = 17.16 cos 8.17t

a) Vmax = 17.16 m/s

b) here u should have 8.17 t = pi = 3.14 approx so that |cos 8.17 t| = 1

=> t = 0.3845 s

c) a(t) = dv/dt = - 17.16 (8.17) sin 8.17t = -140.2 sin 8.17t

d) a(t) max = 140.2 m/s^2

e) 8.17t must equal (3pi/2) in order to get a = + 140.2

=> 8.17 t = 3pi/2 = 4.7
=> t = 0.576 s

f) 2.1 sin [8.17 (1.15)] - 0 = 0.0615 m

Answer 3

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 2.10 cm, and the frequency is 1.30 Hz.

Potential energy =

Answer 4

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