A block with mass m =6.8 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =6.8 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.22 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.9 m/s. The block oscillates on the spring without friction.

Spring constant of the spring: 303.2 N/m

Oscillation frequency: 1.06 Hz

Magnitude of the maximum acceleration of the block: 32.71 m/s^2

1. After t= 0.3 seconds what is the speed of the block??

2. At t= 0.3 seconds what is the magnitude of the net force on the block??

3. Where is the potential energy of the system the greatest? a. highest point of oscillation? b. new equilibrium position of the oscillation? c. at the lowest point of the oscillation?

Explanation / Answer

This has the same numbers as my problem, so I know this method is correct.

(m = mass, g = gravity, k = spring constant, w = angular frequency, f = oscillation frequency, a = acceleration, v = velocity, t = time)

1. At equilibrium the spring constant equals the pull of gravity (net force is zero) so weight is equal to the current distance the string is stretched times the spring constant.

m*g = k * d

2. To solve for oscillation frequency, you must first find the angular frequency.

w = sqrt (k/m)

Then solve for oscillation frequency using

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