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i have derived SHM equation till here $$\sin\left( \frac{\sqrt k}{\sqrt m}\cdot t \right)$$

Now all the solutions (youtube and textbook) I'm looking are just stating intuitively $\frac{k}{m} = \text{constant}$ which is true. But why only $\Omega^2$ and not some other constant? Now if i assume the constant is $\Omega$ then $$\frac{k}{m} = \frac{4π^2}{T^2}$$ while if i just assume some other constant then equation just becomes $\sin(\text{constant}\cdot t)$

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  • $\begingroup$ Derivation must start from the equation of motion, which defines what is what exactly. It appears that you are trying to see why the quoted constant is angular frequency by just staring at the solution. Unless you use physics in the latter approach, arriving at the conclusion is difficult. So, either start from the EOM, or try plotting how an oscillating trajectory looks like and compare coefficient with $\omega$. Also remember the dimensions. $\endgroup$ – physicophilic Jun 20 at 10:14
  • $\begingroup$ @physicophilic i started derivation d^2x/dt^2 +(k/m)*x = 0 and reached 𝑠𝑖𝑛((√𝑘/√𝑚)∗𝑡) $\endgroup$ – Who Jun 20 at 10:16
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    $\begingroup$ I see. My bad to suggest the first point in this case. Plot the solution, and note, that when time increases by $2\pi\sqrt{m/k}$, the function returns to the point you had started from. In other words, the particle with mass m is oscillating with a period $T=2\pi\sqrt{m/k}$. Now, one defines frequency of oscillation $\nu=1/T$. That means the angular frequency must be $\sqrt{k/m}$. $\endgroup$ – physicophilic Jun 20 at 10:23
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    $\begingroup$ are you suggesting that when t=2𝜋√𝑚𝑘 . the function returns to starting point because then t is coming dimensionally incorrect . $\endgroup$ – Who Jun 20 at 11:05
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All you need to do is think about the function $$ \sin( c t ) $$ where $c$ is a constant. For example, you could plot the function on a graph, or find out at what time the function repeats. You will find it repeats when $ct = 2\pi$ and $4\pi,\;6\pi$ etc. So this means that, for this function, the period is $2\pi/c$. This also means that the frequency in cycles per unit time is $c/2\pi$, and the angular frequency (radians per unit time) is $c$. We ordinarily use the symbol $\omega$ or $\Omega$ for angular frequency. Therefore the function is $$ \sin(\Omega t) $$ This means that the mathematical move where you set $$ \frac{\sqrt{k}}{\sqrt{m}} = \Omega $$ is not an arbitrary choice which could have been something else; rather it is a move in which you recognise what role $\sqrt{k/m}$ is playing in the function you have got.

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i started derivation d^2x/dt^2 +(k/m)*x = 0 and reached 𝑠𝑖𝑛((√𝑘/√𝑚)∗𝑡)

But this isn't the entire solution is it? If $X$ is the maximum displacement, then $$x(t) = X\sin(\sqrt{k/m}\cdot t + \phi)$$ is a solution such that the initial position is

$$x(0) = X\sin(\phi)$$

and the initial velocity is

$$\dot{x}(0) = X\sqrt{k/m}\cos(\phi)$$


But why only $\Omega^2$ and not some other constant?

Simple harmonic motion is of the form

$$x(t) = X\sin\left(\frac{2\pi}{T}\cdot t + \phi\right)$$

where $T$ is the period of the motion, i.e., the time over which the motion completes one cycle. The angular frequency is related to the period as so

$$\omega = \frac{2\pi}{T}$$

thus, for the $x(t)$ given above,

$$\omega = \sqrt{k/m}$$

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  • $\begingroup$ 𝑥(𝑡)=𝑋sin(2𝜋𝑇⋅𝑡+𝜙) is the equation i am deriving. how can i already use derived equation to assume that √k/√m.What i am saying is that if we suppose we don't know the equation of shm and deriving it for the first time then there will be no equation to compare and prove that 𝜔=√𝑘/𝑚 $\endgroup$ – Who Jun 20 at 14:37
  • $\begingroup$ @Who, I honestly don't understand what you're looking for after reading your comment above. But, consider this: (1) Given that differential equation, the general solution is the sinusoidal $x(t)$ I give in my answer. (2) The instantaneous angular frequency $\omega$ of a sinusoid $f(t) = \sin\left(\theta(t)\right)$ is defined by $\omega\equiv\dot\theta(t)$. (3) Since $\theta(t) = \sqrt{k/m}\cdot t + \phi$ here, it follows that $\omega = \sqrt{k/m}$, a constant. $\endgroup$ – Hal Hollis Jun 20 at 16:55
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With $\hat i$ a unit vector, a definition of simple harmonic motion might go like this.

The motion of a particle is simple harmonic if

  • the acceleration of the particle, $\vec a$, is proportional to the displacement of the particle from a fixed point, $\vec x$
  • the acceleration of the particle is always directed towards the fixed point

From the first statement $\vec a \propto \vec x \Rightarrow a\,\hat i \propto x \,\hat i$ where $a$ and $x$ are components of acceleration and displacement in the $\hat i$ direction.

From the second statement $\vec a = -c \, \vec x \Rightarrow a\, \hat i = c \, x\,(-\hat i)$ where $c$ ia a constant which must be positive.

How might one ensure that $c$ is positive?
By defining $c=\Omega^2$ where $\Omega$ is a constant. which results in the equation $a = - \Omega^2 \,x$.

Doing this also means that there is a simple relationship between the constant $\Omega$ and the period of the oscillation $T$ which is $T = \frac{2 \pi}{\Omega}$.

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