Simple Harmonic Motion & Pendulum Problems! Need HELP!

1. Feb 14, 2008

dominion_cham

Question 1:
A 0.1 kg block attached to a spring of force constant 16.9 N/m oscillates with an amplitude of 6 cm. Find the maximum speed of the block. Answer in units of m/s.

Question 2:
Find the speed of the block when it is 3 cm from the equilibrium position. Answer in units of m/s.

Question 3:
Find its acceleration at 3 cm from the equilibrium position.

1) Vmax = .06(16.9/.1)^.5
= 0.78 THIS IS CORRECT

2) v = Vmaxsin(pi/3)
= 0.78(pi/3)
= THIS IS WORONG

3) a = .5Vmaxomega
= .5(7.8) (16.9/.1)^.5
= THIS IS WRONG TOO

I tried and I got 2 and 3 wrong :( I ran out of solutions. help please

I was pretty sure one was right.. but it said it was wrong when I entered it :(

Last edited: Feb 14, 2008
2. Feb 14, 2008

rock.freak667

Do you know a formula to find the maximum speed if x=Asin$\omega t$ ?

3. Feb 14, 2008

dominion_cham

Re:

Are you saying that is the speed formula? I am not sure of any.. I feel completely lost!

4. Feb 15, 2008

rock.freak667

The first one is correct.

For the 2nd one, if $x=Asin\omega t$
$\frac{dx}{dt}=v=A\omega cos\omega t$.....(*)

$$sin^2 \omega t+ cos^2 \omega t =1 \Rightarrow cos^2 \omega t=\pm \sqrt{1-sin^2 \omega t}$$

Making (*) become
$$v=\pm \omega A\sqrt{1-sin^2 \omega t}$$

$$v=\pm \omega \sqrt{A^2-A^2sin^2 \omega t}$$

$$v=\pm \sqrt{A^2-x^2}$$

For the third part. What equation do you know connecting acceleration and displacement??

5. Feb 15, 2008

Cynapse

Ok here goes.

The formula for SHM is x = Asin[wt] (w representing omega/angular momentum)
w = sqrt(k/m) Just does :P

So

x = Asin[sqrt(k/m).t]

v=x/t or dx/dt
d(Asin[sqrt(k/m).t])/dt just diffentiates x which is Asin[sqrt(k/m).t], with respect to t
v = dx/dt = sqrt(k/m).Asin[sqrt(k/m).t]

The sin function vaires between 1 and -1,
so v is at a max when sin[sqrt(k/m).t]=1
so you get
vMAX = sqrt(k/m).A

PART 2

For this part you have the same two equations

x = Asin[sqrt(k/m).t]
v = sqrt(k/m).Asin[sqrt(k/m).t]

You are asked to find out what v is at 3cm.
However your v equation depends on t so the first
thing you must do is find at what time t x=3

x = Asin[sqrt(k/m).t]
x/A = sin[sqrt(k/m).t]
sin-1[x/a] = sqrt(k/m).t sin-1 indicates inverse sin
t = (sin-1[x/a])/sqrt(k/m)

with this t you just plug it into your eq for v:
v = sqrt(k/m).Acos[sqrt(k/m).t].

PART 3

a = dv/dt so you just diffrentiate your eq for v again
a = -(k/m).Asin[sqrt(k/m).t]

Now you know what time x=3 so you just put this in the equation for a

If you write these down it will be very clear.

6. Feb 16, 2009

laetitia

i need help with thia problem i seem to be stuck on finding the spring constant here is the problem
a spring is hanging with a 50kg mass on its end. when 20g is added it stretches 7cm more. a)find the spring constant b) if the 20g is removed, what i the new period of motion

7. Feb 24, 2009

skiing4free

a)K=F/x
k=(.02x9.81)/.07
k=2.8Nm^-1
b)f=(1/2$$\pi$$)($$\sqrt{}(k/m)$$
f=(1/2$$\pi$$)($$\sqrt{}(2.8/50)$$
f=0.03766Hz
f=1/t
t=26.6s

Does that help? you should create ur own thread 4 different questions...