At t = 0, a 790 g mass at rest on the end of a horizontal spring (k = 130 N/m)?
At t = 0, a 790 g mass at rest on the end of a horizontal spring (k = 130 N/m) is struck by a hammer, which gives the mass an initial speed of 2.71 m/s.
(a) Determine the period of the motion.
Determine the frequency of the motion.
(b) Determine the amplitude.
(c) Determine the maximum acceleration.
(d) Determine the position as a function of time.
( ___ m ) sin[ ( ___ rad/s)t ]
(e) Determine the total energy.
I was able to figure out the first two parts but couldn't with the rest, any help would be appreciated
- O-360Lv 61 decade agoFavorite Answer
ω = sqrt(k/m) = sqrt(130/.79) = 12.83 rad/sec
f = ω/(2*π) = 2.04 Hz
T = 1/f = 0.49 s
f = 2.04 Hz
The maximum acceleration occurs at the extremes of the oscillation, i.e. when the mass has come to rest. The initial velocity, at the uncompressed length, represents Kinetic Energy (KE):
KE = (1/2)*m*v^2 = (1/2)*.79*(2.71)^2 = 2.9 J
This equals the Potential Energy (PE) at maximum spring compression:
PE = (1/2)*k*x^2 = KE = 2.9 J
x^2 = 2*KE/k
x = sqrt(2*KE/k) = sqrt( 2*2.9/130 ) = 0.211 m
The force at this displacement is:
F = k*x = 130*.211 = 27.5N
F = m*a
a = F/m = 27.5/.79 = 34.8 m/s^2
We determined the amplitude above, and we have the radian frequency from (a):
x(t) = ( .211 m ) sin[ ( 12.83 rad/s)t ]
The total energy is equal to the initial KE, as derived in (c)
E = 2.9 J