At t = 0, a 725 g mass at rest on the end of a horizontal spring (k = 130 N/m) i

Question

At t = 0, a 725 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.
___________________s

Determine the frequency of the motion.
___________________Hz

(b) Determine the amplitude.
___________________m

(c) Determine the maximum acceleration.
___________________m/s2

(d) Determine the position as a function of time.
(___________m ) sin[(____________rad/s)t ]

(e) Determine the total energy.
_______________________J

Explanation / Answer

(a)

= (k/m) = (130/.725) = 13.39 rad/sec

f = /(2*) = 2.1312 Hz

T = 1/f = 0.469 s

(b)

See above

f = 2.1312 Hz

(c)

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*v2 = (1/2)*.725*(2.71)^2 = 2.662 J

This equals the Potential Energy (PE) at maximum spring compression:

PE = (1/2)*k*x2 = KE = 2.662 J

x2 = 2*KE/k

x = (2*KE/k) = ( 2*2.662/130 ) = 0.2023 m

The force at this displacement is:

F = k*x = 130*.2023 = 26.31N

F = m*a

a = F/m = 26.31/.725 = 36.287 m/s2

(d)

We determined the amplitude above, and we have the radian frequency from (a):

x(t) = ( .2023 m ) sin[ ( 13.39 rad/s)t ]

(e)

The total energy is equal to the initial KE, as derived in (c)

E = 2.662 J

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