# Energy and Simple Harmonic Motion.

• sailordragonball
In summary, the block was dropped from a height of 2.49 cm above the compressed spring and the total height of the drop was 4.99 cm. The problem was solved using conservation of energy and taking into account the compression of the spring.
sailordragonball
A vertical spring with a spring constant of 470 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.5 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped?

... I tried using KEo + PEo + SPEo = KEf + PEf + SPEf ... but, I can't get anywhere ...

... I found out that the frequency is 18.05hz and it's relative time is .056 seconds - and that's where I'm stuck!

HELP!

... I found a formula ...

... initial height = (2*mass*gravity)/(spring constant) ...

... but, there's the 2.5cm differential?

... what do I do?

I'd use conservation of energy like you originally suggested. the 2.5 cm compression of the spring comes in there.

I tried that ... but, the lingering the final displacement of 2.5cm is messing me up ... I was wondering ...

(mass*gravity)(initial height + 2.5cm) = .5(spring constant)[(2.5)^2] ...

First -- decide where the potential energy is defined as ZERO. You can put this anywhere... but be consistent.

Then -- look not just at the point where the brick is released, and the point where the brick is compressing the spring, but also think about the intermediate point (when the brick first starts to compress the spring).
Think about that types of energy (PE and or KE) are at each location.

Finally -- make sure the total energy at each location (PE+KE) is equal to the total energy at each other location (PE+KE).

I figured it out:

k = 470(N/m) - given
m = .3kg - given
x = 2.5 cm ... but use 0.025m in the arithmetic
g = 9.8 - given
h = initial height

(mass*gravity)(initial height + displacement) = .5(spring constant)[(displacement)^2]

(.3*9.8)(initial height + .025) = .5(470)[(.025)^2]

2.94(initial height + .025) = .146875

2.94(initial height) + .0735 = .146875

[initial height = .0249m] or [initial height = 2.49cm]

Use initial height = 2.49cm for the remainder of the problem.... now find total height ...

... total height = initial height + displacement ...

... total height = 2.49cm + 2.5cm ...

... total height = 4.99cm

## 1. What is energy?

Energy is the ability to do work or cause change. It exists in many forms, such as kinetic, potential, thermal, electrical, and chemical energy.

## 2. What is simple harmonic motion?

Simple harmonic motion is a type of oscillatory motion in which the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction of the displacement.

## 3. How is energy related to simple harmonic motion?

In simple harmonic motion, energy is constantly being converted between kinetic and potential energy. As the object moves away from the equilibrium position, it gains potential energy and loses kinetic energy. As it moves towards the equilibrium position, the opposite occurs.

## 4. What is the equation for calculating the potential energy in simple harmonic motion?

The equation for potential energy in simple harmonic motion is U = 1/2 kx^2, where k is the spring constant and x is the displacement from the equilibrium position.

## 5. Can energy ever be lost in simple harmonic motion?

In an ideal system, energy is conserved and cannot be lost. However, in real-world systems, energy may be lost due to factors such as friction or air resistance.

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