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A mass m equal to 32 kg is suspended vertically from a spring and, for this reason, it stretches 39.2 cm. Determine the amplitude and period of motion if the mass is released from a point 20 cm above the equilibrium position with an upward velocity of 1 m/s. How many cycles will the mass have completed at the end of 40 s? Suppose g=9.8 m/s²

User Flu
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Answer:

Number of cycles ≈ 14.114

Step-by-step explanation:

To determine the amplitude and period of motion, we can consider the properties of simple harmonic motion (SHM) in the context of a mass-spring system.

Amplitude (A):

The amplitude is the maximum displacement from the equilibrium position. In this case, the mass stretches the spring by 39.2 cm when suspended. Therefore, the amplitude (A) is half of this stretch since the spring's equilibrium position is in the middle of its range of motion:

A = 39.2 cm / 2 = 19.6 cm = 0.196 m

Period (T):

The period is the time taken for one complete oscillation (one cycle) in SHM. The period can be calculated using the following formula:

T = 2π * √(m/k)

Where m is the mass of the object (32 kg) and k is the spring constant.

To find the spring constant (k), we can use Hooke's law:

F = k * x

where F is the force applied by the spring and x is the displacement from the equilibrium position. When the mass is suspended, the force due to the spring is balanced by the force of gravity:

mg = k * A

Solving for k:

k = mg / A

k = (32 kg) * (9.8 m/s²) / 0.196 m

k = 1616 N/m

Now, we can calculate the period (T):

T = 2π * √(m/k)

T = 2π * √(32 kg / 1616 N/m)

T ≈ 2.835 seconds

Number of cycles in 40 seconds:

To find out how many cycles the mass completes in 40 seconds, we can divide the total time by the period:

Number of cycles = Total time / Period

Number of cycles = 40 s / 2.835 s

Number of cycles ≈ 14.114

Since we cannot have a fraction of a cycle, the mass will have completed 14 full cycles at the end of 40 seconds.

User Vahid Ghadiri
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