Question regarding angular frequency of a SHM

Janiceleong26 · Apr 7, 2016

1. Homework Statement

Graph of energy E in mJ against displacement x in cm for a mass on a spring: a downward-opening parabola peaking at about 2.4 mJ at x = 0 and reaching 0 at x = ±1.5 cm, with gridlines and labelled axes from -1.5 to 1.5 cm and 0 to 2.5 mJ.

Text of part (b) of an exam question: the student repeats the investigation with a smaller amplitude, the maximum value of E is now 1.8 mJ, and Fig. 4.2 is to be used to determine the change in amplitude.

Homework Equations

KE=½m(ωa)²

The Attempt at a Solution

So first I did this:
2.4x10^-3= ½ mω²(1.5x10^-2)²
To find mω²=21.33
And substitute that into the KE eqn to find the new amplitude, which is 1.30x10^-2
But I only did that because that was the only way I could think of.
My question is, why is ω constant despite changes in the max KE?

ehild · Apr 7, 2016

Janiceleong26 said:

1. Homework Statement

View attachment 98703
View attachment 98704
Homework Equations

KE=½m(ωa)²

The Attempt at a Solution

So first I did this:
2.4x10^-3= ½ mω²(1.5x10^-2)²
To find mω²=21.33
And substitute that into the KE eqn to find the new amplitude, which is 1.30x10^-2
But I only did that because that was the only way I could think of.
My question is, why is ω constant despite changes in the max KE?

How does the max KE depend on ω and the amplitude?

Physics-Tutor · Apr 7, 2016

Linear velocity will change during the oscillation (x changes)

so the KE

will too:

Equation for kinetic energy in simple harmonic motion: E_k = (1/2) m ω² (x_max² − x²), with E_k and ω written as displayed.

Janiceleong26 · Apr 7, 2016

ehild said:

How does the max KE depend on ω and the amplitude?

Max KE=½m(ωa)²

Janiceleong26 · Apr 7, 2016

@Physics-Tutor Thanks but I'm still confused. How do we know that the angular frequency, ω, is constant?

ehild · Apr 7, 2016

Janiceleong26 said:

Max KE=½m(ωa)²

Changing KE means changing the amplitude if it is the same oscillator. You know that ω=√(k/m). If the spring is the same and the mass of the load is the same, why should the frequency change?

ehild · Apr 7, 2016

Janiceleong26 said:

@Physics-Tutor Thanks but I'm still confused. How do we know that the angular frequency, ω, is constant?

We use the same spring and loading mass.

Physics-Tutor · Apr 7, 2016

Janiceleong26 said:

Thanks but I'm still confused. How do we know that the angular frequency, ω, is constant?

ehild said:

We use the same spring and loading mass.

ehild is correct:

$Formula ω = square root of (k/m), with a small red dot after the fraction, showing angular frequency as the square root of spring constant over mass.$

k is the spring constant, m the mass. These two are constant, so the angular velocity is constant.

Janiceleong26 · Apr 7, 2016

ehild said:

Changing KE means changing the amplitude if it is the same oscillator. You know that ω=√(k/m). If the spring is the same and the mass of the load is the same, why should the frequency change?

Physics-Tutor said:

ehild is correct:
View attachment 98717

k is the spring constant, m the mass. These two are constant, so the angular velocity is constant.

Oh, I got it. Thanks a lot !

Question regarding angular frequency of a SHM

Homework Equations

The Attempt at a Solution

Discussion

Homework Equations

The Attempt at a Solution

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