anonymous
  • anonymous
A particle is executing simple harmonic motion with an amplitude a. when its kinetic energy is equal to its potential energy its distance from the mean position????
Physics
schrodinger
  • schrodinger
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arindameducationusc
  • arindameducationusc
I think A/sqrt(2), I will check it tomorrow nicely and let you know
IrishBoy123
  • IrishBoy123
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IrishBoy123
  • IrishBoy123
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IrishBoy123
  • IrishBoy123
because this is a great question @Michele_Laino @Astrophysics
Astrophysics
  • Astrophysics
Interesting, thanks for the tag @IrishBoy123 Hmm
arindameducationusc
  • arindameducationusc
I am here now, yesterday was late night (my time). Now I will try to give explanation,
anonymous
  • anonymous
First of all thanks for your attention...and an appropriate explanation related to this question would be greatly appreciated
arindameducationusc
  • arindameducationusc
@Shiraz-Khokhar I will be giving detailed explanation shortly, you can be offline for an hour till then... :)
anonymous
  • anonymous
Sure......:)
Michele_Laino
  • Michele_Laino
your motion is a motion of a particle of mass m, which is connected to a spring, whose mass is negligible. That mechanical system can be modeled by the subsequent differential equation: \[m\ddot x = - kx\] with the subsequent initial condition, for position and velocity: \[\begin{gathered} \dot x\left( 0 \right) = 0 \hfill \\ x\left( {\frac{{2\pi }}{\omega }} \right) = a \hfill \\ \end{gathered} \] a is the amplitude of the periodic motion. Now the solution of that differential equation is: \[x\left( t \right) = a\cos \left( {\omega t} \right),\quad {\omega ^2} = \frac{k}{m}\]. The kinetic energy and potential energy are: \[\begin{gathered} T = \frac{1}{2}m\dot x{\left( t \right)^2} = \frac{1}{2}m{a^2}{\omega ^2}{\left\{ {\sin \left( {\omega t} \right)} \right\}^2} \hfill \\ U = \frac{1}{2}kx{\left( t \right)^2} = \frac{1}{2}k{a^2}{\left\{ {\cos \left( {\omega t} \right)} \right\}^2} \hfill \\ \end{gathered} \] then the condition kinetic energy=potential energy, now reads: \[T = U \Rightarrow \frac{1}{2}m{a^2}{\omega ^2}{\left\{ {\sin \left( {\omega t} \right)} \right\}^2} = \frac{1}{2}k{a^2}{\left\{ {\cos \left( {\omega t} \right)} \right\}^2}\] and the solutions are: \[\sin \left( {\omega t} \right) = \pm \cos \left( {\omega t} \right) \Rightarrow \omega t = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\;\frac{{7\pi }}{4}\] Now, if: \[t = \frac{\pi }{{4\omega }}\] then we have: \[x\left( {\frac{\pi }{{4\omega }}} \right) = a\cos \left( {\omega \frac{\pi }{{4\omega }}} \right) = \frac{a}{{\sqrt 2 }}\]
Michele_Laino
  • Michele_Laino
oops... I have made a typo, here are the right initial condition: \[\begin{gathered} x\left( 0 \right) = a \hfill \\ \dot x\left( 0 \right) = 0 \hfill \\ \end{gathered} \]
arindameducationusc
  • arindameducationusc
The total energy of a harmonic oscillator consists of two parts, P.E and K.E. The former being due to its displacement from the mean position and later due to its velocity. Since the position and velocity of the harmonic oscillator are continuously changing, P.E and K.E also change but their sum, i.e, The Total Energy(T.E) must have the same values at all times. Simple harmonic Equation is defined by the equation F=-kx The Work done by the force F during the displacement from x to x+dx is dW =F dx= -kx dx. The work done in the displacement from x=0 to x is W=\[\int\limits_{0}^{x}(-kx)dx=-\frac{ 1 }{ 2 } kx^2\] Let U(x) be the potential energy of the system when the displacement is x. The change in potential energy corresponding to a force is negative of the work done by this force, U(x)-U(0)=-W=1/2kx^2 Let us choose the potential energy to be zero when the particle is at the center of oscillation x=0. Then U(0)=0 and U(x)=1/2kx^2 This expression for potential energy is same as that for a spring. As \[\omega=\sqrt{\frac{ k }{ m}} , k=m \omega ^{2}\] (If you dont know this I will show you in the next post) We can write U(x)=1/2mw^2x^2 The displacement and velocity of a particle executing a simple harmonic motion are given bu x=A sin (wt) and v=Aw cos(wt) //if you dont know why ask me! The potential energy at time t is, Therefor, U=1/2mw^2x^2=1/2mw^2A^2sin^2(wt) =mw^2A^2/4(1-cos2wt) and the kinetic energy at time t is K=1/2mv^2=1/2mA^2w^2cos^2(wt) K=mw^2A^2/4(1+cos2wt) The total mechanical energy at time t is E=U+K = 1/2mw^2A^2 [sin^2(wt)+cos^2(wt)]=1/2mw^2A^2 //if you can't understand the above I will explain in terms of math,(just let me know) This is the Concept of Energy in SHM
arindameducationusc
  • arindameducationusc
Now, P.E=1/2mw^2A^2sin^2(wt) and K.E=1/2mA^2w^2cos^2(wt) for them to be equal, there should be a common value of sin and cos which are equal.... which is sin pi/4= cos pi/4 which gives 1/sqrt(2) hence the amplitude in which K.E=P.E, A/sqrt(2)
IrishBoy123
  • IrishBoy123
|dw:1440142487212:dw|
Astrophysics
  • Astrophysics
Good job @Michele_Laino and @arindameducationusc :)
arindameducationusc
  • arindameducationusc
We are a team. .. Thank you....@Astrophysics and @IrishBoy123 I am quite new in open study but this is awesome..!!
IrishBoy123
  • IrishBoy123
@arindameducationusc i thought it was brilliant that you and @Michele_Laino took different approaches. so much to think about in there but it all fits together perfectly
anonymous
  • anonymous
@arindameducationusc its great but as i m from pre medical section and m not brilliant in mathematics and can you tell me if their is any simpler or straighter method...??
IrishBoy123
  • IrishBoy123
the simplest explanation i can think of still requires a knowledge of a few equations if we stick with a mass-spring system [you mention generic shm but the spring-mass is the complete example], we know that: A] the kinetic energy of the particle KE is \(KE = \frac{1}{2} m v^2\) with mass m and velocity v B] the potential [elastic spring] energy PE is \(PE = \frac{1}{2} k x^2\) where k is the spring constant, and x is extension we also assume that the total amount of energy in the system \(E_T\) is constant, no friction, no air resistance etc.... we can say therefore that \(KE + PE = E_T = const.\) now, when the displacement of the particle is zero, the energy is all in the motion of the mass, it is all KE . and, when the mass is at the extremeties, ie when \(x = \pm A\) -- the energy is all in the form of elastic potential energy of the spring. the particle is at rest as it is about to change direction. thus we can actually calculate \(E_T\). by looking at \(x = A\), we get \(E_T = KE + PE = 0 + \frac{1}{2} k A^2 = \frac{1}{2}k A^2\) turning next to position when when \(PE = KE\), and let's say that occurs at \(x = \bar x\), then \(KE + PE = E_T \implies 2PE = E_T = const.\) \(PE = \frac{1}{2} E_T \) \(\frac{1}{2}k \bar x^2 = \frac{1}{2}. \frac{1}{2}k A^2 \) \( \bar x^2 = \frac{1}{2} A^2 \) \( \bar x = \pm \frac{1}{\sqrt{2}} A \) finally, brief extract from Wiki: simple harmonic motion is a type of periodic motion where the **restoring force is directly proportional to the displacement** and acts in the direction opposite to that of displacement. that is why i say that the frictionless mass-spring system is the complete example of shm. you have Hooke's Law F = kx which does exactly that
anonymous
  • anonymous
Hmmm..thanks buddy...would you tell me why you put 1/2A square in place of Et
anonymous
  • anonymous
IrishBoy123
  • IrishBoy123
there's a 1/2 and a k on the LHS that cancel yes?
anonymous
  • anonymous
Thanks for clearing...i understood previous method but i just wanted to know a quick method because we dont have much time in entry tests...but thankyou so much
Michele_Laino
  • Michele_Laino

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