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aceccardi03
05.08.2019 •
Physics
Ahorizontal spring is lying on a frictionless surface. one end of the spring is attached to a wall, and the other end is connected to a movable object. the spring and object are compressed by 0.060 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 12 rad/s. what is the speed of the object at the instant when the spring is stretched by 0.045 m relative to its unstrained length?
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Ответ:
Velocity = 0.4762 m/s
Explanation:
Given the details for the simple harmonic motion from the question as:
Angular frequency, ω = 12 rad/s
Amplitude, A = 0.060 m
Displacement, y = 0.045 m
The initial Energy = U = (1/2) kA²
where A is the amplitude and k is the spring constant.
The final energy is potential and kinetic energy
K + U = (1/2) mv² + (1/2) kx²
where x is the displacement
m is the mass of the object
v is the speed of the object
Since energy is conservative. So, the final and initial energies are equal as:
(1/2) k A² = (1/2) m v² + (1/2) kx²
Using, ω² = k/m, we get:
Velocity:
Velocity = 0.4762 m/s
Ответ: