At t = 0, a 785g mass at rest on the end of a horizontal spring (k=184N/m) is st

Question

At t = 0, a 785g mass at rest on the end of a horizontal spring (k=184N/m) is struck by a hammer which gives it an initial speed of 2.26 m/s

ANSWER ALL IN APPROPRIATE UNITS

a) Determine the period of motion with appropriate units

b) determine the frequency of motion with appropriate units

c) determine the amplitude

d) determine the maximum acceleration

e) determine the position as a function of time

f) determine the total energy

g) Determine the kinetic energy when x = 0.40A where A is the amplitude.

Explanation / Answer

Mass m = 785 g              = 0.785 kg Spring constant k = 184 N / m Initial speed v = 2.26 m/ s a) The period of motion with appropriate units T = 2[ m / k]                                                                          = 0.41 s b) The frequency of motion with appropriate units f = 1/ T                                                                               = 2.4366 Hz
c) The amplitude A = ? We know v = A                    = A(2f ) From this A = v / (2f )                    = 0.1476 m d) The maximum acceleration a = A 2                                                 = A(2f) 2                                                 = 34.59 m/s 2 e) The position as a function of time x = A cos t                                                          = A cos (2ft)                                                          = 0.1476 cos 15.3 t f) The total energy E = ( 1/ 2) kA 2                                 = 2.004 J

g) The kinetic energy when x = 0.40A is K = ( 1/ 2) k [ A 2 - x 2 ]                                                                   = ( 1/ 2) k [0.84 A 2 ]   Since x = 0.4 A                                                                   = 1.68 J

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