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10 months ago

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in the two cases shown the mass and the spring are identical but the amplitude of the simple harmonic motion is twice as big in

case 2 as in case 1. 1)how are the maximum velocities in the two cases related?

1 answer:

Ira Lisetskai [31]10 months ago

6 0

The maximum velocities in the two cases related are Vmax,2 = 2 Vmax,1

Simple harmonic motion :

In physics, simple harmonic motion is the repeated back-and-forth motion through an equilibrium, or center, position so that the maximum displacement on one side of this position is equal to the maximum displacement on the other. Each whole vibration occurs at the same time interval.

Complete question:

In the two cases shown the mass and the spring are identical but the amplitude of the simple harmonic motion is twice as big in Case 2 as in Case 1.

1)How are the maximum velocities in the two cases related?

Vmax,2 = Vmax,1

Vmax,2 = 2 Vmax,1

Vmax,2 = 4 Vmax,1

To learn more about simple harmonic motion visit: brainly.com/question/28208332

#SPJ4

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kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

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M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

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Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

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v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

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or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

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$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

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$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

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$(0.404V_1)=(3.72)$

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$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

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