Thursday
March 23, 2017

Post a New Question

Posted by on Tuesday, December 7, 2010 at 7:43pm.

At t = 0, a 785 g mass at rest on the end of a horizontal spring (k = 125 N/m) is struck by a hammer, which gives the mass an initial speed of 2.70 m/s.
(a) Determine the period of the motion.
T= .4979201367s
Determine the frequency of the motion.
f= 2.008354 Hz
(b) Determine the amplitude.

(c) Determine the maximum acceleration.

(d) Determine the position as a function of time.

(e) Determine the total energy.

  • Physics - , Tuesday, December 7, 2010 at 7:55pm

    m = 0.785 kg
    k = 125 N/m

    w = sqrt (k/m) = 2 pi f
    T = 1/f

    KE at center of motion = total energy = (1/2) mv^2
    KE = E = (1/2)(m )(2.70)^2
    so
    U at max swing amplitude A = E = (1/2) k A^2
    so
    A^2 = 2 E/k

    max acceleration = w^2 A

    x = A sin w t

    we already did total energy because I chose to find amplitude that way.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question