Consider the following example: A particle oscillates with simple harmonic motion in a straight line. In the first $t$ seconds, it travels a distance $a$. In the next $t$ seconds, it travels an additional $2a$ distance (in the same direction). Find the time period of oscillation. For this question, the correct answer only comes when assuming $x=A\cos \omega t$. When using $x=A\sin \omega t$, the answer does not exist. Why?
123450
Usually every simple harmonic question starts with the line:  The block at equilibrium shifted to a position $X_0$ and released.   I found this from the website:       This equation has a sine in it, and a sine graph starts at zero. Using this equation is like starting your mathematical stopwatch in the middle of a pendulum swing: t = 0 is in the center of the oscillation. If, on the other hand, you replace sine with a cosine, then the equation is still correct; you're just starting to measure time at the maximum displacement instead.   Now if this holds true then most of my question in my textbook in which the mass is  slightly displaced from an equilibrium point uses the sin equation instead of cos. Aren't we measuring motion from the Amplitude point at t=0... Why the discrepancy
68624
in classical physics frequency represents how many cycles in one unit time, but I do not know how we define a frequency for a particle? what does it mean for a particle to have a frequency, the book I read it from just says Energy is proportional to the frequency of the photon but where is frequency?
80468
A damped harmonic oscillator with   $m = 10$kg, $k = 250$N/m, and       $c = 60$kg/s is subject to a driving force given by $F_0\cos(\omega t)$,   where       $F_0 = 48$N.       (a) What value of $\omega$ results in steady-state oscillations with maximum       amplitude?       (b) What is the maximum amplitude? What is the phase shift at the       resonance   Studying for my exam tomorrow, I am kinda of confused to how to even begin this problem. What does it mean by steady-state solution?   I know maximum amplitude occurs when velocity is zero.  The diff equation looks as follows: $$a + \frac{cv}{m} + \frac{kx}{m}$$  and thats all I know. Please do not give me the answer, but please tell me how I can approach this.
109845
I found this excercise in a book ( chapter 4.3):     Show that the function $f(t) = A \sin(at) + B \sin(bt)$ is periodic when   the ratio of $a$ to $b$ is a rational number, and non-periodic if the   ratio is irrational. [Hint: Differentiate twice and take linear   combinations of the result and the original function to get a single   sine wave; use this to get information about possible periods]   I saw general proofs, but I'm interested in using the hint  So, the double derirative:  \begin{array}{l} f''( t) =-a^{2} A\sin( at) -b^{2} B\sin( bt) \end{array}  And what now? Do I just multiply $f(t)$ by $a^2$ and add it to $f''(t)$? Is that a logical thing to do (I mean I haven't had to add a function to its derirative before)  \begin{array}{l} a^{2} f( t) +f''( t) =\left( a^{2} -b^{2}\right) B\sin( bt) \end{array}  Can I conclude anything from this?
49958
I've been pondering the precise mechanism of time dilation for the example of a simple pendulum in two different situations:   The observer and ground are at rest in one frame of reference;  the pendulum is moving at high speed with respect to that frame. The observer is at rest in one frame of reference; the pendulum and the ground together move at high speed with respect to that frame.   user8260 has pointed out that in situation 1, in the pendulum's frame $g$ is greater by $\gamma^2$ compared to $g$ in the observer's frame. Thus in the pendulum's frame the period is less than it is in the observer's frame by a factor of $1/\gamma$, just as one would expect from time dilation.  But what about situation 2? Here, compared to the pendulum frame, the observer sees the pendulum with the same length in a stronger gravitational field, yet observes a longer period. Does the inertial mass of the pendulum change differently than its gravitational mass? Also, does the analysis depend on whether the plane of swing is perpendicular or parallel the direction of motion?
46021
Consider any wave packet describing a free particle (so no potential or other forces acting on it). Then it can be shown that $\Delta p$ does not change in time. However, my question is what happens with $\Delta x$ as we go forward in time? Does it have to increase at all times? Or is there a counter-example where the uncertainty in position is decreasing, if even for a short time period?  My initial guess is $\Delta x$ must always increase, because $p \neq 0$, so that $\Delta p \neq0$ and hence $\Delta v = \frac{\Delta p}{m} \neq 0$. But if there's a spread in velocities, then the wave packet must also spread. Is this logic correct? Or could we have a wave packet where the back of it would move forward faster than the front, and for a certain period until that back end catches up with the front one, it would actually be narrower than at the outset, i.e. reducing $\Delta x$? If yes, how would one describe such a free particle (wave packet)?  So it does seem to me that every wave packet describing a free particle will eventually spread, but the question is whether there can be a time period in its evolution when it is actually becoming narrower.  edit: In particular, if it does not have to increase at all times, can this be shown without appealing to time reversal?
51364
Consider a massless spring system which is hanging vertically. Attach a mass $M$ and set it into simple harmonic motion. Let the period with which the mass oscillates be $T$.   We assume that the spring is massless in most cases. Would taking effect of the non-zero mass of the spring affect the time period ($T$)?
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While studying about the Resonance and Forced Oscillations, I came across a graph in my textbook that is given below:-  Now, the author writes      As the amount of damping increases, the peak shifts to lower frequencies.   Why does this happen? And does this imply that at higher damping levels one cannot achieve a higher amplitude by setting the period of the forced oscillation to be equal to the natural frequency? This seems strange because, the highest amplitude is achieved when the natural frequency is equal to the driving frequency. But, I guess that the rules are somehow different for the damping case.
115767
I am looking for an appropriate derivation of the $(\frac{d}{dt})_{\text{laboratory}} = (\frac{d}{dt})_{\text{rotating}} + \omega \times $ relationship that enables one to calculate all desired quantities in a rotating referece frame. Does anybody know a good way how to understand this transformation?
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In the book 'Calculus the Early Transcendetals' at page 776 (7th edition) they give that the period of a  with length $\text{L}$ that makes a maximum angle $\theta_0$ with the vertical is: $$\text{T}_{\left[\text{s}\right]}=4\sqrt{\frac{\text{L}}{\text{g}}}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1-\sin^2\left(\frac{\theta_0}{2}\right)\sin^2(x)}}\space\text{d}x$$  Questions:  Does this formula work, for any pendulum? How did they get this formula (hint to derive the given formula)?
103685
I solved the Lagrangian of a simple pendulum (with help from online examples as this concept is new to me) and ended with the following: $$\ddot{\theta}+\omega^2\theta=0$$ But in the example I was following, the next line written was this: $$\theta(t) = A\cdot\sin(\omega t + \Phi)$$ And I don't know what steps they took to get there?
112986
I am given an initial x and y position and initial velocity and I was asked to graph the trajectory in 1 second intervals.  This is what I have so far:  If $x_0 = 1, v_{0x} = 70, y_0 = 0, v_{0y} = 80, a_x = 0, a_y = -9.8$, and time will be 0,1,2,3... and so on.  Using these equations on every second you can find the plot will be a bell shaped with the highest point being ~ 325 m at about 600 seconds:   $$ x = x0 + (v_{0x})t + 1/2((a_x)t^2) $$ $$ y = y0 + (v_{0y})t + 1/2((a_y)t^2) $$  Usually in physics, we are taught in perfect condition with no air resistance. But what if there was air resistance? How would it affect this problem? How can i add it to my calculations and see what the difference is?
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I've been taught that in a , for small $x$, $\sin x \approx x$. We then derive the formula for the time period of the pendulum. But I still don't understand the Physics behind it. Also, there's no angle $x$ involved in a spring-mass system, then why do we consider it an SHM only for small amplitudes?
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Suppose you have a physical . It is true that as amplitude increases, the period increases. Can we demonstrate this fact without explicitly finding the period (which is pretty involved and pretty messy) in:    an intuitive fashion,  rigorously?
71949
I recently read in a book that combination of two simple harmonic motions of equal amplitude in perpendicular directions differing in phase by pi/2 is circular motion. I don't seem to understand this because I am not able to figure which two forces in circular motion are acting to cause two different simple harmonic motions.  Any explanation would be thoroughly appreciated
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This is quoted from A.P.French's Vibrations &amp; Waves:  If the driving force is of low frequency  relative to the natural frequency, we would expect the particle to move essentially with the driving force. This is equivalent in saying that $m\dfrac{d^2x}{dt^2}$ in $$m\dfrac{d^2x}{dt^2} + kx = F_0\cos\omega t$$ plays small role compared to the term $kx$. The amplitude is controlled by spring constant. On the other hand, at frequencies very large compared to the natural frequency, the opposite situation holds. The $kx$ becomes small compared to $m\dfrac{d^2x}{dt^2}$. In this case, we expect a relatively smaller amplitude of oscillation than the above case.  $\bullet$ Why does $kx$ become prominent in the first case &amp; not in the second case? $\bullet$ What is the physical reason behind that the first case has greater amplitude &amp; the second one smaller?
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The concept of velocity is by definition the movement divided by the interval of time between initial position and final position.  If $f(t)$ is the position of a particle at time $t$; the velocity in the interval $[t_0;t_1]$ is $\dfrac{f(t_1)-f(t_0)}{t_1-t_0}$  The problem is that in a single instant there is no movement and the time is not changed; so no velocity.  I can consider $\lim_{t_1 \to t_0} \dfrac{f(t_1)-f(t_0)}{t_1-t_0}$, but mathematically it is only the limit of average velocity function and doesn't represent velocity at instant $t_0$  What are your views about this problem ?