Calculating Oscillatory Motion Parameters for a Spring-Block System

[MissPenguins]

Homework Statement

A 0.7 kg block attached to a spring of force
constant 13.4 N/m oscillates with an ampli-
tude of 20 cm.
Find the maximum speed of the block.
Answer in units of m/s.
(part 2 of 4) 10.0 points
Find the speed of the block when it is 10 cm
from the equilibrium position.
Answer in units of m/s.
(part 3 of 4) 10.0 points
Find its acceleration at 10 cm from the equi-
librium position.
Answer in units of m/s2.
014 (part 4 of 4) 10.0 points
Find the time it takes the block to move from
x = 0 to x = 10 cm.
Answer in units of s.



I already answered part 1, and 2. I got .875 in the first part, .7578 for the second part. But I don't know how to do part 3 and 4.

Homework Equations

For part 3, I tried using v = \pm\sqrt{}k/m(A²-x₂)

The Attempt at a Solution

I tried to find the derivative of the above equation for acceleration. I don't think I did the derivative right. Can someone please help?

 

[ehild]

You certainly learned Hook's law about how much force does a stretched spring exert on a body attached to its end. And you know that force = mass times acceleration. So how much is the force of the spring when it is stretched by 10 cm from its equilibrium position?

As for the time it takes the block to move from x=0 to x=10: This motion has the time dependence x = A sin (wt). You know w from the spring constant and mass. The amplitude is given: 20 cm. At t = 0, x=0. Find the argument of the sine when x/A =10/20. Divide by w.

ehild

 

[MissPenguins]

The Force isn't given for in order to use F = ma. So I used a = -w²Acos(wt+theta). w²=k/m. I did all those and I got -70.1599. Is it right? Does it make sense?

For the second part, I used x(t) = A cos(wt), so I calculated (0.1 m)/(0.2 m) = 0.5 m. cos inverse, then I found the t to be 1.047198. Is it correct?Thanks for your help.

 

[ehild]

The Force isn't given for in order to use F = ma. So I used a = -w²Acos(wt+theta). w²=k/m. I did all those and I got -70.1599. Is it right? Does it make sense? .

You do know the force of a spring: it is F = -k(x-x0). x-x0 = 0.1 m is the deviation from the equilibrium position, k is the spring constant, k=13.4 N/m. How much is the force?

You can use also the formula a = -w²Acos(wt+theta), but I do not see how did you got 70.16. In what unit is it?

For the second part, I used x(t) = A cos(wt), so I calculated (0.1 m)/(0.2 m) = 0.5 m. cos inverse, then I found the t to be 1.047198. Is it correct?


Thanks for your help.

No it is not correct. You have to get the elapsed time from the equilibrium position, that is x=0, to x=0.1 m. At what time is x=0 ?

ehild

 

[MissPenguins]

You do know the force of a spring: it is F = -k(x-x0). x-x0 = 0.1 m is the deviation from the equilibrium position, k is the spring constant, k=13.4 N/m. How much is the force?

You can use also the formula a = -w²Acos(wt+theta), but I do not see how did you got 70.16. In what unit is it?



No it is not correct. You have to get the elapsed time from the equilibrium position, that is x=0, to x=0.1 m. At what time is x=0 ?

ehild

Alright, for the first part, I did F = -k(x-x₀) = (- 13.4 N/m)(-0.1 m) =1.34 N.
F = ma => F/m = a, 1.34 N/ 0.7 kg = 1.914 m/s². Is that correct?

I still don't get the other part. Thanks. :)

 

[ehild]

You assumed that the position of block is x=A cos(wt), but it can be taken x = A sin (wt), as well. It is the same motion, only the starting time is different. If x = 0.2 sin(wt) the block is at equilibrium position at t=0. What is wt when it is at x=0.1 next? See picture.

ehild

 

[MissPenguins]

You assumed that the position of block is x=A cos(wt), but it can be taken x = A sin (wt), as well. It is the same motion, only the starting time is different. If x = 0.2 sin(wt) the block is at equilibrium position at t=0. What is wt when it is at x=0.1 next? See picture.

ehild

wt is at 0.5 when x = 0.1.

 

[Bearbull24.5]

I have this same problem, different values. I did x=Asin(wt) .035=.07sin(1.632t). .035/.07=sin(1.632t). .5/1.632=sint. t=sin^-1(.306186). Got an answer of 17.829 s which was wrong

 

[ehild]

MissPenguins:
Sin(wt)=0.5--->wt=sin^-1(0.5)= pi/6=0.5236.
w=4.375 s^-1 ---> t=0.12 s.


Bearbull24.5: You can not "factor out" a multiplayer from the argument of a function. So 0.5=sin(1.632t) ---->1632t = sin^-1(0.5)=pi/6.


ehild

 

[Bearbull24.5]

So I get sin^-1(.5)=30 divide the by w (1.63299) and I get the wrong answer (18.37117)

 

[ehild]

So I get sin^-1(.5)=30 divide the by w (1.63299) and I get the wrong answer (18.37117)

Write out the units. Now you got the time t in degree/rad *s. Find the angle in radians.


ehild

 

[MissPenguins]

MissPenguins:
Sin(wt)=0.5--->wt=sin^-1(0.5)= pi/6=0.5236.
w=4.375 s^-1 ---> t=0.12 s. Bearbull24.5: You can not "factor out" a multiplayer from the argument of a function. So 0.5=sin(1.632t) ---->1632t = sin^-1(0.5)=pi/6. ehild

Can you please tell me where did you get 4.375 s from? thanks.

Nvm, I got it w = sqrt of k/m. Hehe, I got the right answer too, thank you very much.

 

[ehild]

!.375 s^-1 is the angular frequency w.

w = sqrt(spring force/mass)

ehild