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Amplitude and Period

1. Homework Statement
A massless spring of spring constant k = 74 N/m is hanging from the ceiling. a 490 g mass is hooked onto the unstretched spring and allowed to drop.

A) Find the amplitude
B) Find the period of the resulting motion

2. Homework Equations
f = 1/T
F=-kx
x(t) = Acos (omega) t
omega = sqrt( k / m )
x(t) = Acos(omega * t + phi )
omega = sqrt ( g / L )
T = 2*PI * sqrt ( L / g )




3. The Attempt at a Solution

I'm lost in this chapter and have no idea where to start, i was curious if someone could help step me through this problem, would help me a ton.

Thanks!!
 

Answers and Replies

for part b)

omega = sqrt (k / m)

freq = omega / (2PI)

T = 1/freq

i cant figure out how to find amplitude
 
alphysicist
Homework Helper
2,238
1
Hi nightshade123,

Try using conservation of energy at the highest and lowest points.
 
at the highest and lowest points v = 0 and Potential Energy is at its max and kinetic energy is 0

i know i have this eqns

E = U + K

U = 1/2 K A^2 (cos(omega t ))^2
K = 1/2 K A^2 (sin(omega t ))^2
E = 1/2 K A^2
 
1/2kx + 1/2mv = 1/2kA

kx + mv = kA

v = 0

kx = kA

x=A

F = -k * x

x = - F/k

F = mg

x = (mg)/k

x = A
 
335
0
The amplitude is the maximum extension in the spring.
Calculate omega using omega = (k/m)^1/2.
Now calculate extension using hook's law.
Now for max extension = kx^2=mv^2
Calculate velocity.Now use v= omega*Amplitude to get amplitude.
Time period = 2pi/omega.
 
you dont have to do all that work A = x work it out yourself. A = .065 m thx for the help tho
 
alphysicist
Homework Helper
2,238
1
nightshade123,

1/2kx + 1/2mv = 1/2kA

kx + mv = kA

v = 0

kx = kA

x=A
This part isn't quite right. In the formula for the energies the x, v, and A have to be squared. Also, you get x=A but it doesn't tell you what either of them is. The problem is in setting up the equation like this (where x=0 at the equilibrium point).

Instead you could try setting up the equation with x=0 at the spring's unstretched point which would then explicitly include gravity in the energy equation. Letting x=0 and h=0 at the higest point, and setting enegy to be conserved between the higest and lowest points gives:

[tex]
\begin{align}
\frac{1}{2}k x_i^2 + mgh_i &=\frac{1}{2}k x_f^2 + mgh_f\nonumber\\
0+0 &= \frac{1}{2} k x^2 + m g (-x)\nonumber
\end{align}
[/tex]

Solving this for the (nonzero) value of x would give the entire range of motion from highest to lowest point; the amplitude would be half of that.




F = -k * x

x = - F/k

F = mg

x = (mg)/k

x = A
This alternative method looks good. The distance from the highest point to the equilibrium position is the amplitude and so the x value you found here equals A.
 

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