Simple Harmonic Motion & Pendulum Problems Need HELP

In summary,The block oscillates with an amplitude of 6 cm and has a maximum speed of 0.78 m/s when 3 cm from equilibrium. When 20 g is added to the block, it stretches by 7 cm and the speed decreases to 0.78 m/s.
  • #1
dominion_cham
2
0
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.
Answer in units of m/s2.1) 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:
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  • #2
Do you know a formula to find the maximum speed if x=Asin[itex]\omega t[/itex] ?
 
  • #3


Are you saying that is the speed formula? I am not sure of any.. I feel completely lost!
 
  • #4
The first one is correct.

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

[tex]sin^2 \omega t+ cos^2 \omega t =1 \Rightarrow cos^2 \omega t=\pm \sqrt{1-sin^2 \omega t}[/tex]

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

[tex]v=\pm \omega \sqrt{A^2-A^2sin^2 \omega t}[/tex]

[tex]v=\pm \sqrt{A^2-x^2}[/tex]

For the third part. What equation do you know connecting acceleration and displacement??
 
  • #5
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
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
laetitia said:
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

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

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

Related to Simple Harmonic Motion & Pendulum Problems Need HELP

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion where an object oscillates back and forth around a stable equilibrium position. It occurs when the restoring force on the object is directly proportional to its displacement from the equilibrium position, and the motion follows a sinusoidal pattern.

2. How is a pendulum related to simple harmonic motion?

A pendulum consists of a mass attached to a fixed point by a string or rod. When the mass is pulled to one side and released, it will swing back and forth, exhibiting simple harmonic motion. The restoring force in this case is gravity, which is directly proportional to the displacement of the mass from its equilibrium position.

3. What factors affect the period of a pendulum?

The period of a pendulum is affected by the length of the pendulum, the mass of the pendulum, and the acceleration due to gravity. The longer the length of the pendulum, the longer the period, and the greater the mass of the pendulum, the longer the period. The period is also affected by the strength of the gravitational field, which can vary depending on the location.

4. How do you calculate the period of a pendulum?

The period of a pendulum can be calculated using the equation T = 2π√(l/g), where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity. This equation assumes small angles of oscillation, which is commonly the case for most pendulums.

5. What is the relationship between the amplitude and energy of a simple harmonic motion?

The amplitude of a simple harmonic motion is directly proportional to the energy of the system. This means that as the amplitude increases, so does the energy. However, the frequency and period of the motion remain constant, regardless of the amplitude. This is because the energy in a simple harmonic motion is dependent on the square of the amplitude.

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