At time t=0, a 0.75 kg mass at rest on the end of a horizontal spring is struck

Question

At time t=0, a 0.75 kg mass at rest on the end of a horizontal spring is struck very quickly by a hammer, which give it an initial speed of 2.76 m/sec, but no appreciable displacement during the strike. The spring is of a type that can be both compressed and stretched from its equilibrium position. The same 0.75 kg mass, if hung vertically from the spring, would stretch it 0.06 meters. (Use g=10.0 m/s^2)

a. What is the value of the spring constant? (use units)

b. What is the frequency of the motion?

c. What is the amplitude of the motion?

d. What is the total energy of the motion?

e. Give all locations where the mass would have a speed of 1.50 m/sec.

Explanation / Answer

Mass m = 0.75 kg

Initial speed v = 2.76 m/ s

a) Spring constant, k = (0.75 x 10)/0.06 = 125 N/m

b)The period of motion with appropriate units T = 2[ m / k] = 0.486 s

The frequency of motion with appropriate units f = 1/ T = 2.05 Hz

c) The amplitude A = ?

We know v = A = A(2f )

From this A = v / (2f ) = 0.214 m

d) The total energy E = ( 1/ 2) kA 2 = 2.856 J

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