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If there is no external force with explicit time dependence, then the harmonic oscillator contains no explicit time dependence. Then the system has time translation symmetry, i.e. the result can only depend on the difference $T =t_b-t_a$, not on $t_a$ and $t_b$ individually.


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The answers currently posted are ignoring a few important details so I'm going to give my own. I may rehash some things already said. To make everything absolutely clear I write here a complete derivation of the forced damped oscillator with emphasis on the role of the $Q$ factor. Basic equations Consider the equation of motion of a forced, damped harmonic ...


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If an object is accelerating upwards at a rate of $a$ m/s$^2$, then the gravitational force felt by this object is effectively, $$ g_{eff}=g+a $$ where $g\sim9.8$ m/s$^2$ is the canonical gravitational acceleration we all know and love. In your particular case, the common equation of motion for a pendulum, $$ \frac{d^2\theta}{dt^2}=mg\sin\theta $$ replaced ...


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You can think of the continuos formalism as being the limiting case of the discrete-momentum one: if the momentum is taken as a discrete variable (which amounts to constraining the particles to be in some finite volume $V$) the fourier expansion of the (real, scalar) field is: $$ \tag{1} \varphi(x) = \sum_{\textbf{k}} \frac{1}{\sqrt{2V \omega_{\textbf{k}}}} ...


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He uses a non-inetrial reference frame, where the pendulum's bob is at rest, so there are two fictitious forces: the $$\text{centrifugal force} = mL(\frac{d\theta}{dt})^2$$, that acts radially and out, and the $$\text{Euler force} = $mL\frac{d^2\theta}{dt^2}$$ ,that acts tangentially. Then the tension, $T$, along $L$, the gravity vertically, a tangential ...


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You can resolve this using the following FBD Tension in Pink Damping Forces (from linear and rotational dampers) in Red Inertial Forces in Gray. (Note: $\ddot{\theta}=\dot{\omega}=\alpha$)


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You actually do not need to solve the equation at all in order to determine that, whatever the answer, it must be the same for A and B. The only difference between A and B is the direction of k1. However if you look at $F=-kx$ is always a restoring force whose magnitude is proportional to the magnitude of the displacement from equilibrium. So the mass ...


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simple. the box is confined in space by two springs. there are only two forces acting on the box, $F_1$ and $F_2$, exerted by those two springs. Well, the two forces must balance each other out by Newton's third law, or the law of action-reaction pairs. so we get $$ F_1 = -F_2 = -k_1 x_1 = k_2 (-x_2) --> k_1 x_1 = k_2 x_2 $$ (note that $x_1$ and $x_2$ ...


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Take a look at the junction of the two springs. If you displace the block and hold it there, the junction will also be at rest. At the junction, two forces are pulling it, $k_1 x_1$ to the left and $k_2 x_2$ to the right. Since the junction is at rest, the two forces must balance. Therefore, $$k_1 x_1 = k_2 x_2$$ EDIT 1: Another calculus-based ...


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Well it depends on the context of your question. If you're being introduced to General Relativity, then you're just going to assume, in the spirit of the equivalence principle, that gravity and the acceleration cannot be told apart from the pendulum's standpoint, so the acceleration is obviously $a+g$. If you need to do it from first principles in a ...


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It is true in both cases and an even more general statement can be made. For a potential of the form $V(r)=\alpha r^n$, the expectation value of the kinetic an potential energy is related by $$ 2 \langle T \rangle = n \langle V \rangle. $$ For the case of the harmonic oscillator, $n=2$. See virial theorem for details.


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The fact that you can eliminate the $\sin$ term tells you that its coefficient $[-\omega^2 B_{2}-2\beta\omega B_{1}+\omega_{0}^{2}B_{2}]$ must be zero. This gives you another relation between $B_{1}$ and $B_{2}$, and with two equations and two unknowns, you can solve cleanly for $B_{1}$ and $B_{2}$.


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Because (as you say) the ODE is linear we have that if $\phi_i$ are all valid solutions then so is $$\sum_ia_i\phi_i$$ for any real $a_i$. You can convince yourself of this by substituting the sum into the ODE and showing it is satisfied assuming the $\phi_i$ are solutions. We use the sin and cosine functions for our decomposition because they happen to ...



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