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# 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.

.49 s

Determine the frequency of the motion.

2.04 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

I was able to figure out the first two parts but couldn't with the rest, any help would be appreciated

### 1 Answer

- O-360Lv 61 decade agoFavorite Answer
(a)

ω = sqrt(k/m) = sqrt(130/.79) = 12.83 rad/sec

f = ω/(2*π) = 2.04 Hz

T = 1/f = 0.49 s

(b)

f = 2.04 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*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

(d)

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

x(t) = ( .211 m ) sin[ ( 12.83 rad/s)t ]

(e)

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

E = 2.9 J

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