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dimulka [17.4K]

3 years ago

11

A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi

tion and released, it oscillates with a period of 2 seconds.Four English majors are discussing what would happen to the period of oscillation if the cart was displaced 12 cm from its rest position instead of 6 cm and again released.With which, if any, of these students do you agree?

1 answer:

Marianna [84]3 years ago

8 0

Answer:

Time period of horizontal Oscillation = T = 2 $\pi$ $\sqrt{\frac{m}{k} }$

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation = T = 2 $\pi$ $\sqrt{\frac{m}{k} }$

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

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We can solve this part of the problem by using the law of conservation of energy. In fact:

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Equalizing the two expressions, we can solve to find A, the amplitude:

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