oscillation
Exercise - converting variables 1
A mass $m=5kg$ is connected to a spring with $k = 20N/m$. Find the angular frequency, frequency, and period.
Use $\omega = \sqrt{\frac{k}{m}}$ and $\omega = \frac{2\pi}{T} = 2\pi f$.
Solution
$$
\begin{eqnarray}
\omega &=& \sqrt{\frac{k}{m}} \\
&=& \sqrt{\frac{20}{5}} = 2rad/s
\end{eqnarray}
$$
Using $\omega = \frac{2\pi}{T} = 2\pi f$ we can find the other variables:
$$
\begin{eqnarray}
f &=& \frac{\omega}{2\pi} \\
&=& \frac{2}{2\pi} = 0.32Hz \\
T &=& \frac{2\pi}{\omega} \\
&=& \pi = 3.14s
\end{eqnarray}
$$
angular frequency =
2
0.1
$rad/s$
frequency =
0.32
0.02
$Hz$
period =
3.14
0.02
$s$
Feedback will appear here.
oscillation
Exercise - converting variables 2
A mass $m=5kg$ is tied to a string $2m$ long to form a simple pendulum. Find the angular frequency, frequency, and period.
Use $\omega = \sqrt{\frac{g}{l}}$ and $\omega = \frac{2\pi}{T} = 2\pi f$.
Solution
$$
\begin{eqnarray}
\omega &=& \sqrt{\frac{g}{l}} \\
&=& \sqrt{\frac{9.8}{2}} = 2.21rad/s
\end{eqnarray}
$$
Using $\omega = \frac{2\pi}{T} = 2\pi f$ we can find the other variables:
$$
\begin{eqnarray}
f &=& \frac{\omega}{2\pi} \\
&=& \frac{2.21}{2\pi} = 0.35Hz \\
T &=& \frac{2\pi}{\omega} \\
&=& \frac{2\pi}{2.21}= 2.84s
\end{eqnarray}
$$
angular frequency =
2.21
0.03
$rad/s$
frequency =
0.35
0.03
$Hz$
period =
2.84
0.03
$s$
Feedback will appear here.
oscillation
Example - measuring acceleration due to gravity
In an experiment, it is observed that a $2m$-long pendulum oscillates $65$ times in $160s$. Use this to deduce the acceleration due to gravity.
Solution
Each oscillation takes:
$$
\begin{eqnarray}
T &=& \frac{160}{65} = 2.46s\end{eqnarray}
$$
This is the period of the pendulum. From the period, we could find $\omega$ using $\omega=\frac{2\pi}{T}$:
$$
\begin{eqnarray}
\omega &=& \frac{2\pi}{T} = \frac{2\pi}{2.46} = 2.55rad/s
\end{eqnarray}
$$
Since $\omega = \sqrt{\frac{g}{l}}$, we have:
$$
\begin{eqnarray}
\omega &=& \sqrt{\frac{g}{l}}\\
\Rightarrow g &=& \omega^2 l = 2(2.55)^2\\
&=& 13.03m/s^2
\end{eqnarray}
$$
Therefore according to this experimental result, $g = 13.03m/s^2$. It is a bit off from $9.8m/s^2$, but it is probably due to some experimental error, or maybe the experiment was performed on a different planet.
oscillation
Exercise - measuring the spring constant 1
In an experiment, it is obsersed that a mass $m=3kg$ tied to a spring oscillates $100$ times in $486s$. Find the period of the oscillation.
If you are tempted to look up a fancy equation to answer this question, you are already down the wrong path. Just think about the following questions:
If you eat $2$ candies in $50s$, how long does it take you to each one candy? It is $\frac{50s}{2}=25s$.
If you eat $10$ candies in $50s$, how long does it take you to each one candy? It is $\frac{50s}{10}=5s$.
If you eat $100$ candies in $50s$, how long does it take you to each one candy? It is $\frac{50s}{100}=0.5s$.
If you eat $100$ candies in $486s$, how long does it take you to each one candy? (you get the idea)
Now if the mass has $100$ oscillations in $486s$, how long does it take to complete one oscillation? This is the definition of period.
Solution
Open the hint for explanation if you do not understand this calculation below. The time it takes for one oscillation (i.e. the period) is:
$$
\begin{eqnarray}
T &=& \frac{486s}{100} = 4.86s
\end{eqnarray}
$$
period =
4.86
0.01
Select unit:
$rad/s$
$Hz$
$s$
$N/m$
2
Feedback will appear here.
oscillation
Exercise - measuring the spring constant 2
Continue from the previous problem. Deduce the spring constant of the spring.
Use $\omega = \frac{2\pi}{T}$ to find $\omega$, then $\omega = \sqrt{\frac{k}{m}}$ to find $k$.
Solution
From $\omega = \frac{2\pi}{T}$:
$$
\begin{eqnarray}
\omega &=& \frac{2\pi}{T} = 1.29 rad/s
\end{eqnarray}
$$
Using $\omega = \sqrt{\frac{k}{m}}$, we have:
$$
\begin{eqnarray}
k &=& m \omega^2 = 3(1.29)^2 = 5.01N/m
\end{eqnarray}
$$
spring constant =
5.01
0.2
Select unit:
$rad/s$
$Hz$
$s$
$N/m$
3
Feedback will appear here.
oscillation || calculus
Example - displacement, velocity, and acceleration of oscillations 1
Given $x=A\cos(\omega t + \phi)$, where $A=2m$, $\omega=3rad/s$, and $\phi= \pi$, find the velocity and acceleration of the oscillator at $t=1.5s$.
Solution
We will use $v = \dot{x}$ and $a = \dot{v}$ to find the velocity and acceleration:
$$
\begin{eqnarray}
v &=& \dot{x} = \frac{d}{dt} A\cos(\omega t + \phi) \\
&=& A\frac{d}{dt} \cos(\omega t + \phi) \\
&=& - A \omega \sin (\omega t + \phi)\\
a &=& \dot{v} = \frac{d}{dt} \left( - A \omega \sin (\omega t + \phi) \right) \\
&=& - A \omega \frac{d}{dt} \sin (\omega t + \phi) \\
&=& (- A \omega) \left(\omega \cos (\omega t + \phi) \right) \\
&=& -A \omega^2 \cos (\omega t + \phi)
\end{eqnarray}
$$
Putting in the numbers we get:
$$
\begin{eqnarray}
v(1.5s) &=& - (2) (3) \sin ((3)(1.5) + 3.14) = -5.87m/s \\
a(1.5s) &=& -(2) (3^2) \cos ((3)(1.5) + 3.14) = -3.79m/s^2
\end{eqnarray}
$$
Actually, we did not really have to differentiate twice to get $a$ (although we did it to get some math practice). The trick is to note $a = \dot{v} = \ddot{x} = -\omega^2 x$, where we used the SHO equation ($\ddot{x} = -\omega^2 x$). Therefore, we could have just computed $x=A\cos(\omega t + \phi)$ at $t=1.5s$ and just multiply it by $-\omega^2$. You could check that it gives exactly the same result.
oscillation || calculus
Example - displacement, velocity, and acceleration of oscillations 2
Given $x=A\cos(\omega t + \phi)$, where $A=2m$, $\omega=3rad/s$, and $\phi= \pi$, find the maximum velocity and acceleration of the oscillator.
Solution
From the last example, we have:
$$
\begin{eqnarray}
v &=& - A \omega \sin (\omega t + \phi)\\
a &=& -A \omega^2 \cos (\omega t + \phi)
\end{eqnarray}
$$
Because both $\sin$ and $\cos$ functions are bounded within $-1$ and $+1$, the general rule for finding the maximum value of such functions are given by:
$$
\begin{eqnarray}
\big( M \cos \theta \big)_{max} &=& |M| \\
\big( M \sin \theta \big)_{max} &=& |M|
\end{eqnarray}
$$
Therefore we have:
$$
\begin{eqnarray}
v_{max} &=& |- A \omega | = A\omega = 6m/s \\
a_{max} &=& |-A \omega^2| = A \omega^2 = 18m/s^2
\end{eqnarray}
$$