Need Help!-->>Physics (Oscillations)

[Tweety-Angie]

I would be really glad if anyone could help me out with these two questions.

1. A particle is oscillating in simple harmonic motion with frequency 50Hz and amplitude 15mm. Calculate the speed when the displacement from the equilibrium position is 12mm.

2. a spring stretches by 85mm when a mass of 50g is hung from it. The spring is then stretched a further distance of 15mm from the equilibrium position, and the mass is released at time t=0. Calculate: (Doubt1: What will be the equilibrium state here? when x=0 or x=85mm?? )
(a) the spring constant, (My doubt2: which length should we put here 85mm or 15mm? :ugeek: Why? :? )
(b) the amplitude of the oscillations, (Doubt3: I think the amplitude is 15mm. Is it correct? :?: )
(c) the period, :?
(d) the displacement at time t=0.20s.

Thanks in advance, hope to have response soon...

 

[DragonCub]

For the first question, differentiation, I think, is required.
The function can be interpreted using trigonometry: d = 15 sin (...)
Here d refers to the displacement in mm.
The thing in the bracket can be converted from T into radians (not degrees, or the data would not be compatible).
The frequency is 50 Hz, that is, 0.02 second = 2π Rad, so 1 second = 100Π Rad.
Thus d = 15 sin (100π * T)

Now is the time to employ differentiation!!
dd/dt = 15 * 100π * cos (100π * T)
dd/dt is just the velocity.

When d =12 mm, 15 sin (100π * T) = 12
sin (100π * T) = 4/5
So we can get cos (100π * T) = 3/5 (No need to calculate the value of T!)
Hence dd/dt = 1500π * (3/5) = 900π mm/s
That is the velocity, so the speed.

 

[WellWIshER]

wat the flip!!!!

thts too hard core.!!

the ans is rong too.
da ans i suppose is 2.8 M/S

 

[DragonCub]

Well it is very close then.
900π mm/s ≈ 2827.4 mm/s ≈ 2.8m/s

 

[WellWIshER]

Oh SO ITS 900pi!!

why dont we stick to da formula v=w sqt A^2-x^2