A mass m is attached to an ideal massless spring. When this system is set in motion with amplitude a, it has a period t. What is
1 answer:
The period will be the same if the amplitude of the motion is increased to 2a
What is an Amplitude?
Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.
Here,
mass m is attached to the spring.
mass attached = m
time period = t
We know that,
The time period for the spring is calculated with the equation:
Where k is the spring constant
Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.
From the equation, we can say,
Time period of the spring is independent of the amplitude.
Hence,
Increasing the amplitude does not affect the period of the mass and spring system.
Learn more about time period here:
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This question involves the concepts of the law of conservation of momentum and velocity.
The velocity of the eight ball is "5.7 m/s".
According to the law of conservation of momentum:
where,
m₁ = mass of number three ball = 5 g
m₂ = mass of the eight ball = 6 g
u₁ = velocity of the number three ball = 3 m/s
u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)
v₁ = final velocity of the three number ball = - 5 m/s
v₂ = final velocity of the eight ball = ?
Therefore,
(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)
<u>v₂ = 5.7 m/s</u>
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Answer:
H = 6.93 m
Explanation:
given data
velocity v = 35 m/s
horizontal component Vx = 33 m/s
solution
we get here maximum height so first we get vertical component here that is express as
Vy = .........................1
put here value
Vy =
Vy = 11.66 m/s
and
now we get height
H = .............................2
put here value
H =
H = 6.93 m