Calculating Wave Amplitude and Frequency

[Edel Crine]

Homework Statement

A restoring force of unknown magnitude is exerted on a object
that oscillates with a period of 0.50 s.
When the object is in an evacuated container, the motion is simple harmonic motion, with an amplitude of 0.10 m.
When air is allowed into the container, the amplitude decreases by 2.0% with each cycle of the oscillation.
(a) What is the amplitude after 25 cycles?
(b) What fraction of the initial energy is left 6.3 s after air is admitted?

Relevant Equations

A\left(t\right)=A_{0}e^{-\frac{t}{2τ}}
w_{d}=w\sqrt{1-\frac{\frac{b^{2}}{m^{2}}}{4w^{2}}}

My attempts were these,
a) 2.0% / cycle * 25 cycles = 50%
So, I got half of the first amplitude which is 0.5 m (seems not right though...)

b) w=2pi/T , so put 0.5 at T, I got w=12.6 cycle/sec
12.6 cycle / sec * 6.3 sec = 79.2 cycles and it is obviously not right to me...

May I get some help from you all...? Thank you so much...!

 

[PeroK]

If the amplitude is $0.1m$ how much does it reduce by in one cycle? If the amplitude is $0.08m$ how much does it reduce by in one cycle?

What is a percentage reduction?

 

[Edel Crine]

If the amplitude is $0.1m$ how much does it reduce by in one cycle? If the amplitude is $0.08m$ how much does it reduce by in one cycle?

What is a percentage reduction?

For the first one, it would be reduced by 0.002m / cycle and second would be 0.0016 m / cycle?

 

[Edel Crine]

If the amplitude is $0.1m$ how much does it reduce by in one cycle? If the amplitude is $0.08m$ how much does it reduce by in one cycle?

What is a percentage reduction?

Oopps I worte it wrong, my first answer for a) was 0.05m, not 0.5m

 

[PeroK]

For the first one, it would be reduced by 0.002m / cycle and second would be 0.0016 m / cycle?

But in part a) you used the same reduction for all cycles. You took $25 \times 0.002m$.

Hint: have you ever heard of exponential decay?

 

[haruspex]

the amplitude decreases by 2.0% with each cycle of the oscillation.

I would interpret that as the amplitude reducing by 2% of what it was in the preceding cycle, not by 2% of its initial amplitude.

 

[Edel Crine]

But in part a) you've used the same reduction for all cycles.

Hint: have you ever heard of exponential decay?

yes like A(t)=Ae^(-t/2τ)

 

[PeroK]

yes like A(t)=Ae^(-t/2τ)

Can you see how that applies here?

 

[Edel Crine]

I would interpret that as the amplitude reducing by 2% of what it was in the preceding cycle, not by 2% of its initial amplitude.

Woo yeah, I've misunderstood the problem since now..

 

[Edel Crine]

Can you see how that applies here?

I know initial amplitude and T for both period and 6.3 sec, but I am not sure about tau, τ...

 

[PeroK]

I know initial amplitude and T for both period and 6.3 sec, but I am not sure about tau, τ...

You can calculate $\tau$ from the given decay rate:

Homework Statement:: A restoring force of unknown magnitude is exerted on a object
that oscillates with a period of 0.50 s.

When air is allowed into the container, the amplitude decreases by 2.0% with each cycle of the oscillation.

 

[Edel Crine]

You can calculate $\tau$ from the given decay rate:

I just searched, so we can use the equation, τ=1/λ which is inverse of decay rate?

 

[Edel Crine]

You can calculate $\tau$ from the given decay rate:

Then, 1/0.02 = 50, so the equation would be A(t) = Ae^-t/100 ?

 

[PeroK]

I just searched, so we can use the equation, τ=1/λ which is inverse of decay rate?

You need to plug the numbers into get $\tau$. You know that if $t = 0.5s$, then the amplitude is $0.98A_0$.

Note: for part a) you are given a whole number of cycles: $25$. There is a quick way to calculate the amplitude: hint think of compound interest. But, in part b) the time is $6.3s$, which is not a whole number of cycles. That suggested to me that you were expected to think of exponential decay. That allows you to calculate things for any time, even when it's not a whole number of cycles.

 

[PeroK]

b) w=2pi/T , so put 0.5 at T, I got w=12.6 cycle/sec
12.6 cycle / sec * 6.3 sec = 79.2 cycles and it is obviously not right to me...

May I get some help from you all...? Thank you so much...!

Note that if $T = 0.5s$, then that is $2$ cycles per second (by definition). $\omega$ is the angular frequency, which is not cycles per second.

 

[Edel Crine]

Note that if $T = 0.5s$, then that is $2$ cycles per second (by definition). $\omega$ is the angular frequency, which is not cycles per second.

for τ, I got 12.37 sec and 0.06m for a)...!

 

[PeroK]

That answer looks right.

For part b) what is the relationship between energy and amplitude for SHM?

 

[Edel Crine]

That answer looks right.

For part b) what is the relationship between energy and amplitude for SHM?

E=1/2 mw^2A^2 ...?

 

[PeroK]

E=1/2 mw^2A^2 ...?

Yes, proportional to the square of the amplitude. You can get the energy from the amplitude. And you know how the amplitude decays. So ...

Hint: remember that $mw^2 = k$ is just a constant.

 

[Edel Crine]

Yes, proportional to the square of the amplitude. You can get the energy from the amplitude. And you know how the amplitude decays. So ...

Hint: remember that $mw^2 = k$ is just a constant.

So, it would be just Eo/Ed which is simply A0^2/Ad^2...?

 

[PeroK]

So, it would be just Eo/Ed which is simply A0^2/Ad^2...?

What's $d$?

 

[Edel Crine]

What's $d$?

Oh it's for the symbol of damped amplitude.

 

[PeroK]

Oh it's for the symbol of damped amplitude.

So, what's your answer to part b)?

 

[Edel Crine]

So, what's your answer to part b)?

About 60.1% from the original...?

 

[PeroK]

Looks good.

 

[Edel Crine]

Looks good.

Thank you so much...! I really really appreciate!