- #1
alexvong1995
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I have learned in 1st year that the under-damped simple harmonic motion can be described by the differential equation [tex] m \frac {d^2 x} {dt^2} + b \frac {dx} {dt} + kx = 0 [/tex] where [itex] m [/itex] is the mass, [itex] b [/itex] is the constant of linear drag and [itex] k [/itex] is the spring constant
But the derivation is skipped and only the solution is given, which is [tex] x(t) = x_m e^{-\frac {bt} {2m}} \cos(\omega' t + \phi)[/tex] Recently, I have learned Laplace transform on YT, so I try to come up with my own derivation, which is as followed.
[tex] m \frac {d^2 x} {dt^2} + b \frac {dx} {dt} + kx = 0 [/tex]
[tex] m s^2 X(s) - m s x(0) - m x'(0) + b s X(s) - b x(0) + k X(s) = 0 [/tex]
Let the initial displacement [itex] = x(0) = x_0 [/itex], initial velocity [itex] = x'(0) = 0 [/itex]
[tex] m s^2 X(s) - m s x_0 + b s X(s) - b x_0 + k X(s) = 0 [/tex]
[tex] X(s) (m s^2 + b s + k) = x_0 (m s + b) [/tex]
[tex] X(s) = x_0 \frac {m s + b} {m s^2 + b s + k} [/tex]
[tex] X(s) = x_0 \frac {s + \frac {b} {m}} {s^2 + \frac {b} {m} s + \frac {k} {m}} [/tex]
Complete the square,
[tex] X(s) = x_0 \frac {s + \frac {b} {m}} { (s^2 + \frac {b} {m} s + \frac {b^2} {4 m^2}) + \frac {k} {m} - \frac {b^2} {4 m^2}} [/tex]
[tex] X(s) = x_0 \frac { (s + \frac {b} {2m}) + \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } [/tex]
[tex] X(s) = x_0 \left [ \frac { s + \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } + \frac { \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } \right ] [/tex]
Apply the First Shifting Theorem,
[tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \sqrt{\frac {k} {m} - \frac {b^2} {4 m^2}} t \right ) + x_0 \frac {b} {2m} \frac {1} { \sqrt { \frac {k} {m} - \frac {b^2} {4 m^2} } } e^{- \frac {b} {2m} t } \sin \left ( \sqrt{\frac {k} {m} - \frac {b^2} {4 m^2}} t \right ) [/tex]
[tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + x_0 \frac {b} {2m} \frac {2 m} { \sqrt{4 k m - b^2} } e^{- \frac {b} {2m} t } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) [/tex]
[tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + x_0 \frac {b} { \sqrt {4 k m - b^2} } e^{- \frac {b} {2m} t } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) [/tex]
To convert the answer into the form [itex] x_m e^{-\frac {bt} {2m}} \cos(\omega' t + \phi) [/itex], we must compute the hypotenuse of the triangle with side [itex] 1 [/itex] and [itex] \frac {b} { \sqrt {4 k m - b^2} } [/itex]
[tex] \sqrt{ 1^2 + \frac {b^2} {4 k m - b^2} } [/tex]
[tex] = \sqrt{ \frac {4 k m - b^2 + b^2} {4 k m - b^2} } [/tex]
[tex] = 2 \sqrt{ \frac {k m} {4 k m - b^2} } [/tex]
[tex] x(t) = 2 x_0 e^{- \frac {b} {2m} t } \sqrt{ \frac {k m} {4 k m - b^2} } \left [ \frac {1} { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) \right ] [/tex]
If [itex] \cos \phi = \frac {1} { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } [/itex], then [itex] \sin \phi = \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } [/itex]
By [itex] \cos(A) \cos(B) + \sin(A) \sin(B) = \cos(A-B) [/itex],
[tex] x(t) = 2 x_0 e^{- \frac {b} {2m} t } \sqrt{ \frac {k m} {4 k m - b^2} } \cos \left [ \frac { \sqrt{4 k m - b^2} } {2 m} t - \arccos \left ( \frac {1} {2} \sqrt{ \frac {4 k m - b^2} {k m} } \right ) \right ] [/tex]
Is the derivation correct?
I am pretty confident with [itex] \omega' = \frac { \sqrt{4 k m - b^2} } {2 m} [/itex]
However, I am not so sure about the amplitude and the phase. Is the following true?
[tex] x_m = 2 x_0 \sqrt{ \frac {k m} {4 k m - b^2} } [/tex]
and
[tex] \phi = \arccos \left ( \frac {1} {2} \sqrt{ \frac {4 k m - b^2} {k m} } \right ) = \arcsin \left ( \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \right ) = \arcsin \left ( \frac {b} {2 \sqrt{k m} } \right ) [/tex]
Finally, is there an easier way to come up to a solution? Doing these seems to be very exhausting and prone to careless mistakes. Also, I want to learn about over-damping and critical damping. Isn't the same differential equation apply? What are their differences? Thanks for reading my question! I know it is long!
But the derivation is skipped and only the solution is given, which is [tex] x(t) = x_m e^{-\frac {bt} {2m}} \cos(\omega' t + \phi)[/tex] Recently, I have learned Laplace transform on YT, so I try to come up with my own derivation, which is as followed.
[tex] m \frac {d^2 x} {dt^2} + b \frac {dx} {dt} + kx = 0 [/tex]
[tex] m s^2 X(s) - m s x(0) - m x'(0) + b s X(s) - b x(0) + k X(s) = 0 [/tex]
Let the initial displacement [itex] = x(0) = x_0 [/itex], initial velocity [itex] = x'(0) = 0 [/itex]
[tex] m s^2 X(s) - m s x_0 + b s X(s) - b x_0 + k X(s) = 0 [/tex]
[tex] X(s) (m s^2 + b s + k) = x_0 (m s + b) [/tex]
[tex] X(s) = x_0 \frac {m s + b} {m s^2 + b s + k} [/tex]
[tex] X(s) = x_0 \frac {s + \frac {b} {m}} {s^2 + \frac {b} {m} s + \frac {k} {m}} [/tex]
Complete the square,
[tex] X(s) = x_0 \frac {s + \frac {b} {m}} { (s^2 + \frac {b} {m} s + \frac {b^2} {4 m^2}) + \frac {k} {m} - \frac {b^2} {4 m^2}} [/tex]
[tex] X(s) = x_0 \frac { (s + \frac {b} {2m}) + \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } [/tex]
[tex] X(s) = x_0 \left [ \frac { s + \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } + \frac { \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } \right ] [/tex]
Apply the First Shifting Theorem,
[tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \sqrt{\frac {k} {m} - \frac {b^2} {4 m^2}} t \right ) + x_0 \frac {b} {2m} \frac {1} { \sqrt { \frac {k} {m} - \frac {b^2} {4 m^2} } } e^{- \frac {b} {2m} t } \sin \left ( \sqrt{\frac {k} {m} - \frac {b^2} {4 m^2}} t \right ) [/tex]
[tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + x_0 \frac {b} {2m} \frac {2 m} { \sqrt{4 k m - b^2} } e^{- \frac {b} {2m} t } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) [/tex]
[tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + x_0 \frac {b} { \sqrt {4 k m - b^2} } e^{- \frac {b} {2m} t } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) [/tex]
To convert the answer into the form [itex] x_m e^{-\frac {bt} {2m}} \cos(\omega' t + \phi) [/itex], we must compute the hypotenuse of the triangle with side [itex] 1 [/itex] and [itex] \frac {b} { \sqrt {4 k m - b^2} } [/itex]
[tex] \sqrt{ 1^2 + \frac {b^2} {4 k m - b^2} } [/tex]
[tex] = \sqrt{ \frac {4 k m - b^2 + b^2} {4 k m - b^2} } [/tex]
[tex] = 2 \sqrt{ \frac {k m} {4 k m - b^2} } [/tex]
[tex] x(t) = 2 x_0 e^{- \frac {b} {2m} t } \sqrt{ \frac {k m} {4 k m - b^2} } \left [ \frac {1} { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) \right ] [/tex]
If [itex] \cos \phi = \frac {1} { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } [/itex], then [itex] \sin \phi = \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } [/itex]
By [itex] \cos(A) \cos(B) + \sin(A) \sin(B) = \cos(A-B) [/itex],
[tex] x(t) = 2 x_0 e^{- \frac {b} {2m} t } \sqrt{ \frac {k m} {4 k m - b^2} } \cos \left [ \frac { \sqrt{4 k m - b^2} } {2 m} t - \arccos \left ( \frac {1} {2} \sqrt{ \frac {4 k m - b^2} {k m} } \right ) \right ] [/tex]
Is the derivation correct?
I am pretty confident with [itex] \omega' = \frac { \sqrt{4 k m - b^2} } {2 m} [/itex]
However, I am not so sure about the amplitude and the phase. Is the following true?
[tex] x_m = 2 x_0 \sqrt{ \frac {k m} {4 k m - b^2} } [/tex]
and
[tex] \phi = \arccos \left ( \frac {1} {2} \sqrt{ \frac {4 k m - b^2} {k m} } \right ) = \arcsin \left ( \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \right ) = \arcsin \left ( \frac {b} {2 \sqrt{k m} } \right ) [/tex]
Finally, is there an easier way to come up to a solution? Doing these seems to be very exhausting and prone to careless mistakes. Also, I want to learn about over-damping and critical damping. Isn't the same differential equation apply? What are their differences? Thanks for reading my question! I know it is long!