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2 years ago

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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?.

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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Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

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To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

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Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

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$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

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$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

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$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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