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RoseWind [281]
2 years ago
12

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

the period if the amplitude of the motion is increased to 2a?.
Physics
1 answer:
Luden [163]2 years ago
7 0

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:

T = 2\pi \sqrt{\frac{m}{k} }

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:

<u>brainly.com/question/13834772</u>

#SPJ4

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A bike and rider together have a mass of 60 kg. If the bike and rider have an acceleration of 2.0 m/s^2, what is the force on th
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The net force on the bike and the rider is 120 N

Explanation:

We can solve this problem by applying Newton's second law of motion, which states that:

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A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0 when the han
slava [35]

There are mistakes in the question as the unit of speed and height is not mention here.The correct question is here

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0m/s when the hand is 1.80m above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Answer:

t=3.37s

Explanation:

Given Data

As we have taken hand at origin and positive upward

So given data are

y_{i}=0m\\y_{f}=-1.80m\\v_{i}=16.0m/s\\a=g=9.8m/s^{2}

To find

time taken by the ball before it hits the ground

Solution

By using the common kinematic equation

y_{f}=y_{i}+v_{i}t+0.5at^{2}

Put the given values and find for t

So

-1.80=0+16.0t+(0.5*(-9.8)t^{2} )\\-1.80=16.0t-4.9t^{2}\\ 4.9t^{2}-16.0t-1.80=0

Apply quadratic formula to solve for t

t=\frac{-(-16.0)+\sqrt{(-16)^{2}+4(4.9)(-1.80)} }{2(4.9)}\\ t=3.37s

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