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A 0.60kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.30m. what is the velocity when it passes the equilibrium point.

2007-05-19 06:58:49 · 5 answers · asked by Anonymous in Science & Mathematics Physics

5 answers

You don't really need the amount of mass to solve this problem; the amplitude and frequency of vibration is enough. The velocity of something undergoing simple harmonic motion as it swings through the point of equilibrium is

v = 2 * pi * f * A

where

v = velocity as it goes through the point of equilibrium,

f = oscillation frequency, and

A = amplitude of the vibration.

Most people don't know calculus, even those in physics courses, unless they are math, science or engineering students far enough along in their studies to have had some. If you happen to be one that has had some calculus, it would be easy enough to differentiate the expression

x = A sin (wt)

where
x is the position of the mass as a function of time,

w is the angular frequency, which is equal to 2 * pi * f,

and t is time,

to derive the expression for velocity, which is

v = w * A * cos (wt)

and then picking a value of t where the weight is swinging through its point of equilibrium. In this example, of those times is t = 0. At that time, cos (wt) = 1. Converting w back into frequency gives the expression already quoted for the velocity through the equilibrium point, namely,

v = 2 * pi * f * A
= 2 * pi * 3 cycles/sec * 0.30 m
= 1.8 * pi m/sec
or about 5.65 m/sec

2007-05-19 08:43:29 · answer #1 · answered by devilsadvocate1728 6 · 0 0

Energy = 1/2 m vmax^2 = 1/2 D xmax^2

D = 2 pi m f

==> vmax = 2 pi f xmax =
2 pi 3/s 0.3m = 5.65m/s

2007-05-19 14:46:45 · answer #2 · answered by wolf 6 · 0 0

Well, position vs. time is sinusoidal. You have the amplitude and frequency terms. The time derivative is velocity. What's its amplitude?

2007-05-19 15:39:56 · answer #3 · answered by Dr. R 7 · 0 0

use SHM, let me get my physics paper all the equationsa re at the back brb
back
at the equilibrium point v is maximum,
v=A2pi.fcos(A2pi.f.t)
so when v is max cos(A2pi.f.t) = 1, ( thats the biggest it could be)
so:
vmax=A2pi.f

0.3*2*pi*3=1.8pi=5.65(3sf)

5.65 m/s

2007-05-19 14:07:52 · answer #4 · answered by Anonymous · 1 0

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2007-05-19 14:09:31 · answer #5 · answered by Anonymous · 0 2

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