Please help with this pendulum problem, ASAP. Thanks!?

Question

I really need the answer to this pendulum problem ASAP. Thank you so much!

A pendulum of length L has a bob of mass M at the end of it. The bob is attached to a spring that has a force constant k. When the bob is directly below the pendulum support, the spring is unstressed. Derive an expression for the period of this oscillating system for small-amplitude vibrations (assume there is no displacement from the horizontal). Suppose that M = 2.80 kg and L is such that in the absence of the spring the period is 2.90 s. What is the force constant k if the period of the oscillating system is 1.45 s?

Answer 1

Displacement of the bob x.

Angle of inclination α = x/L.

Force acting on the bob:

F(x) = -kx - Mg(x/L) = - (k+Mg/L)x



Dynamic equation:

Ma = F(x)

M d²x/dt² + (k+Mg/L)x = 0

d²x/dt² + (k/M+g/L)x = 0

d²x/dt² + ω²x = 0, where

ω² = (k/M+g/L)



****************** *************

T = 2π/ω = 2π/√(k/M+g/L)

****************** *************



Now:

2π/T = ω = √(k/M+g/L)

(2π/T)² = (k/M+g/L)

∆(2π/T)² = k/M





Answer:

k = (2π)²M ∆(1/T)² = 39.43 N/m

Answer 2

T = 2pi / angular frequency



for the period



There are obviously a couple of forces here to consider. You will need to consider these before attempting to do any of the problem, I think



Hooke's law for the spring (H)

The weight of the Bob (w = mg)

Answer 3

If A is the backside part of the pendulum, from the regulation of conservation of capability Kinetic capability at A = ability capability at 0.a hundred and fifty m above A Or a million/2 mV^2 = mgh Or V = sqrt (2gh) = sqrt (2 x 9.8 x 0.a hundred and fifty) = a million.715 m/s

Answer 4

How can the spring be unstressed when the bob is directly below the pendulum support?