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3 years ago

When a mass on a spring is at maximum displacement, what quantity is at a minimum?

Physics

1 answer:

Vlada [557]3 years ago

5 0

Answer:

The velocity

Explanation:

For a mass-spring system, the total mechanical energy is constant during the motion. The total mechanical energy is sum of the elastic potential energy, U, and the kinetic energy, K:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$

where

k is the spring constant

x is the displacement

m is the mass

v is the velocity

Since E must remain constant, we see that when x increases, v decreases, and vice-versa. Therefore, when x (the displacement) is at maximum, v (the velocity) is at minimum (more precisely, it is zero).

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According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

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Where:

$G$ is the Gravitational Constant

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If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

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$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

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If Planet X had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be 2T.

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