Stiffness constant of a spring.

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1. Sep 24, 2015

upender singh

1:question
A weight W is suspended from a rigid support by a hard spring with stiffness constant k . The spring is enclosed in a hard plastic sleeve, which prevents horizontal motion, but allows vertical oscillations. A simple pendulum of length l with a bob of
mass m (mg<<W) is suspended from the weight W and is set oscillating in a horizontal line with a small amplitude. After some time has passed, the weight W is observed to be oscillating up and down with a large amplitude (but not hitting the sleeve). It follows that the stiffness constant k ‬ must be:

1. k=4w/l
2. k=2w/l
3. k=w/l
4. k=w/2l
ans:1

2:relevant equations
w^2=k/m

3:attempt
Please hint me where to start. I tried balancing the forces but that brought me nowhere.

Last edited: Sep 24, 2015
2. Sep 25, 2015

lightgrav

the pendulum mass has vertical acceleration component (v^2/r) as it swings.
The spring must pull the mass upward to provide that acceleration, so the mass pulls the spring downward.
If the timing of the pendulum pulls is appropriate, the spring's oscillations will increase in Amplitude (resonance).
(the pendulum's downward pull does positive Work to the spring if the spring is moving downward, meanwhile).

3. Sep 26, 2015

upender singh

thanks for the help lightgrav,

please reply if i am missing something

let Ω = oscillation frequency of spring mass system alone
let ω= oscillation frequency of the simple pendulum,
then mv^2/r =ml(Asinωt)^2 = mlA^2(1-cos2ωt)/2 , A being some constant
which says that driving force has a frequency= 2ω
equating both for resonance,
k*g/W = 4*g/l
k=4W/l

I assumed that mg wont contribute to the oscillations since mgcosθ≅mg for small θ.