For a spring oscillating in simple harmonic motion, at what point will the velocity of
1 answer:
Answer:
equilibrium position.
Explanation:
In simple harmonic motion , velocity v(t) is given by,
v(t) = -ω A sin(ωt + φ)
where
ω = angular velocity of the corresponding circular motion
A = amplitude
t = time
φ = the initial angle of the corresponding circular motion when the motion begin.
v (t) get maximized when sin value is maximized , i.e. sin =1
The particle has maximum speed when it passes through the equilibrium position.
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Answer:
a) 23 m/s
Explanation:
- Assuming no external forces acting during the collision, total momentum must be conserved, as follows:
- The initial momentum p₀, can be written as follows:
- The final momentum pf, can be written as follows:
- Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:
- This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.