Is there is any change in time period if it is done inside accelerating lift?

AI Thread Summary
The discussion revolves around the effects of acceleration on the time period of simple harmonic motion (SHM) in a lift. It addresses three scenarios: a floating wooden block in water, a spring, and a pendulum, with varying conclusions on whether their time periods change due to the lift's acceleration. The consensus is that while the equilibrium position may shift, the time period for the block remains valid under specific conditions. Additionally, the conversation highlights the potential for errors in textbook answers, emphasizing the importance of understanding the underlying physics. Ultimately, the analysis confirms the correctness of the initial calculations despite the complexities introduced by the lift's motion.
vkash
Messages
316
Reaction score
1

Homework Statement



Q1> there is a cube like jar of water kept at the surface of a lift which is moving with an acceleration of a in upward direction. jar is filled with water. There is a cubical block of wood floating on it's surface. It is slightly pressed and released. find the time period of the Simple harmonic motion.(viscosity of liquid is zero)

Q2> A spring is attached to ceiling in lift will it's time period(of oscillations) change with acceletation of life(I think no).

Q3>A pendulum is attached to ceiling will it's time period vary with acceleration.(I think yes T=2πsqrt(l/g))

After all i come to conclusion that "it is mixed one. I mean if it is inside a lift it may or may not have same time period as it had"

Homework Equations



Archimedes principle and Simple equations of SHM.

The Attempt at a Solution



let me say after slight push it's mean position is displaced by x0.
After a distance x below x0 net force acting on the wooden block is dAx(g+a).
here d is density A is base area of block and g is acceleration due to gravity.
F=dAx(g+a) =mω2x (m is mass of block)
from here T(time period)=2πdqrt(m/[dA(g+a)] ).

IF i am wrong somewhere then please alert me.

(recently updated to firefox 9. it is not responding to spelling mistake so if i have done any mistake then please ignore that)

thanks for reading...
waiting for ur valuable reply..
 
Last edited:
Physics news on Phys.org
I'm assuming the acceleration of the lift is meant to be constant with time? In this case, I think you have got all the answers correct :) One spelling mistake though - you've written dqrt, but I think you mean sqrt (i.e. square root).
 
BruceW said:
I'm assuming the acceleration of the lift is meant to be constant with time? In this case, I think you have got all the answers correct :) One spelling mistake though - you've written dqrt, but I think you mean sqrt (i.e. square root).

thnaks BruceW for answering.
Finally the answer behind the book is wrong and i am correct.
 
about question 1>, there is also the force of gravity on the wooden block. But you still have the right answer, because this only affects where the point of equilibrium is. Also, the amplitude of oscillation must be small enough that the block doesn't jump out of the water. As long as this is true, then you've got the right answer for the period.

Textbooks sometimes do have wrong answers (usually due to error when the person wrote it). They probably checked it through a few times, but a few mistakes always seem to slip through the net. It shows that you understand the material if you can spot the errors :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top