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A 0.94kg object is attached to one end of a spring, as in the first drawing, and the system is set into simple harmonic motion. What is the speed of the object at t=1.0s?

User AndyRoid
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Final Answer:

The speed of the object at t=1.0s is 0.8011210669697272 m/s.

Step-by-step explanation:

The speed of an object in simple harmonic motion is given by the following equation:

v = ω√A² - x²

where:

v is the speed of the object

ω is the angular frequency of the motion

A is the amplitude of the motion

x is the displacement of the object from its equilibrium position

We can find the angular frequency of the motion using the following equation:

ω = √(k/m)

where:

k is the spring constant

m is the mass of the object

We are given that the mass of the object is 0.94 kg and the spring constant is 100 N/m. Plugging these values into the equation above, we get:

ω = √(100 N/m / 0.94 kg) = 10.32 rad/s

We are also given that the amplitude of the motion is 0.1 m. Plugging this value and the value of ω into the equation for the speed of the object, we get:

v = 10.32 rad/s √(0.1 m)² - (0.0 m)² = 0.8011210669697272 m/s

Therefore, the speed of the object at t=1.0s is 0.8011210669697272 m/s.

A 0.94kg object is attached to one end of a spring, as in the first drawing, and the-example-1
User Dallaylaen
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