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a 300-g mass at the end of a Hookean spring vibrates up and down in such a way that it is 2.0 cm above the tabletop at its lowest point and 16cm above at its highest point. Its period is 4.0 s. Determine (a) the amplitude of vibration, (b) the spring constant, (c) the speed and acceleration of the mass when it is 9 cm above the table top, (d) the speed and acceleration of the mass when it is

User Karlee
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Final answer:

The amplitude of vibration is 7 cm, the spring constant is 1.178 N/cm, the speed of the mass when it is 9 cm above the tabletop is 9.75 cm/s, and the acceleration is -35.03 cm/s^2.

Step-by-step explanation:

The amplitude of vibration for a mass attached to a Hookean spring can be calculated using the formula:

Amplitude = (Highest point - Lowest point)/2

For this problem, the amplitude is (16 cm - 2 cm)/2 = 7 cm.

The spring constant can be determined using the formula:

Spring constant (k) = (2π/period)^2 * mass

Using the given period of 4.0 s and mass of 300 g (0.3 kg), the spring constant is calculated as:

(2π/4.0)^2 * 0.3 = 1.178 N/cm

The speed and acceleration of the mass when it is 9 cm above the tabletop can be determined using the equation:

v = √(k/m) * √(A^2 - x^2)

a = -k * x / m

where v is speed, k is the spring constant, m is the mass, A is the amplitude, and x is the displacement from the equilibrium position. Plugging in the values, we get:

Speed = √(1.178/0.3) * √(7^2 - 9^2) = 9.75 cm/s

Acceleration = -1.178 * 9 / 0.3 = -35.03 cm/s^2

The speed and acceleration when the mass is at its highest or lowest point will be zero.

User Subvertallchris
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