Simple harmonic motion - damping introduced

[Jozefina Gramatikova]

Homework Statement

Homework Equations

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The Attempt at a Solution

I can solve everything but d). Please help

 

[Charles Link]

The damping coefficient is $b$ in your formula. Do you know how to solve $e^{-\frac{b}{2m}t}=\frac{1}{2}$ for $t$ ? . $\\$ Edit: A google of the topic calls $\gamma=\frac{b}{2m}$ the damping coefficient, but it is really a choice of terminology. By the units they give you, they are giving you $b$.

 

[Jozefina Gramatikova]

The damping coefficient is $b$ in your formula. Do you know how to solve $e^{-\frac{b}{2m}t}=\frac{1}{2}$ for $t$ ? .

I don't know how much is x(t)

 

[Charles Link]

I don't know how much is x(t)

The sinusoidal oscillation is assumed to happen at a much higher frequency with small damping, so that the period of the oscillation $T$ is quite short, and you don't need to consider the term $\cos(\omega t )$. The amplitude is $A e^{- \frac{b}{2m} t}$.

 

[Jozefina Gramatikova]

The sinusoidal oscillation is assumed to happen at a much higher frequency with small damping, so that the period of the oscillation $T$ is quite short, and you don't need to consider the term $\cos(\omega t )$. The amplitude is $A e^{- \frac{b}{2m} t}$.

Ok, thank you and what about x(t)

 

[Charles Link]

Your (first equation) that you posted in part 2 of the OP for $x(t)$ is correct. (You didn't list an equation for $\omega'$ though). And your second equation, which I think is a damping ratio number is really not of prime interest here. It would help if you would state what the parameter is. I had to go googling for the second equation you posted, to see if it was correct.$\\$ When they say, how long does it take for the amplitude to "halve" that means $A e^{-\frac{b}{2m} t}=\frac{1}{2} Ae^{-\frac{b}{2m} 0}$, so that $e^{-\frac{b}{2m} t}=\frac{1}{2}$.

 

[Jozefina Gramatikova]

Oh, I am sorry I didn't think an equation for w' would be helpful here. Thank you very much for your help!

 

[Charles Link]

Now, the next step is take the natural log of both sides of this last equation in order to solve for $t$. (It may be worthwhile for me to mention that, because I'm not sure how advanced you may be).

 

[Jozefina Gramatikova]

Now, the next step is take the natural log of both sides of this last equation in order to solve for $t$. (It may be worthwhile for me to mention that, because I'm not sure how advanced you may be).

Yeah, I know how to proceed from here $A e^{-\frac{b}{2m} t}=\frac{1}{2} Ae^{-\frac{b}{2m} 0}$,. I got t=3.036s. I hope that this is correct

 

[Charles Link]

Suggest you use $\ln{2}=.693$ and you get $t=3.05$ seconds. (I see you must have used $\ln{2}=.690$).

 

[Jozefina Gramatikova]

That's true. I did plug 0.69. Thank you!