A 4 kg mass is connected to a spring (k = 730 N/m) platform which undergoes simp

Question

A 4 kg mass is connected to a spring (k = 730 N/m) platform which undergoes simple harmonic motion in the vertical direction. The amplitude of the oscillation is 5 cm. At t=0 the box is 5 cm above the equilibrium position.
(a) What is the period of oscillation?
s

(b) What is the angular frequency of the box?
rad/s

(c) What is the velocity of the box at the equilibrium position?
m/s

(d) What is the total energy of the 4 kg block?
J

(e) What is acceleration (magnitude and direction) of the box after 5 s (assume mass is at +A at t = 0 s.)?
m/s2

Explanation / Answer

the equation of the motion is :

(mrac{d^2h}{dt^2}+kh=0)

where h is the height of the box above the equillibriium position.

the solution is (h=Acos(omega t+phi))

where is angular frequency and is given by, (omega=sqrt{rac{k}{m}})

(b) thus, (omega=sqrt{730/4}=13.5 s^{-1})

where A is amplitude and is phase.

amplitude is 5cm=0.05 m, thus A=0.05 m

at t=0, h=0.05 m , thus,

(0.05=0.05cos(phi))

which gives, (cosphi=1)

thus, (phi=0)

thus, (h=0.05cos(omeaga t))

(a) time period is (T=rac{2pi}{omega}=0.465 s)

(c) The velocity is given by

(v=rac{dh}{dt}=-0.05omega sin(omega t))

at equillibrium position v is maximum, which happens for (sin(omega t)=-1)

thus, (v = 0.05*omega=0.05*13.5=0.675 m/s)

(d) total energy is :

(E=rac{1}{2}k A^2=0.5*730*0.05^2=0.915 J)

(e) The acceleration is given by,

(a=rac{dv}{dt}=-5omega^2 cos(omega t))

thus at t=5 s,

(a=-0.05*(13.5)^2 cos(13.5*5)=-3.49 m/s^2)

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