====== M4. Spring pendulum ======

===== - Introduction =====
In this experiment, you have to quantify some of the factors affecting the period of oscillation of a spring pendulum and determine its spring constant. A spring pendulum is created by attaching a mass $m$ on the lower end of a vertical spring and let the mass go through a simple harmonic motion due to the gravitational force exerted on the mass by the earth.

===== - Mathematical formalism =====
{{ https://upload.wikimedia.org/wikipedia/commons/2/22/Spring_pendulum.gif?nolink }}

The motion of a spring pendulum is governed by the interplay of the elastic restoring force of the spring and the gravitational force of the earth. Both the elastic and the gravitational forces are exerted on the mass attached to the lower end of the spring.

The differential equation of motion for a spring in simple harmonic motion is given by the linear acceleration

$$ \frac{d^2x}{dt^2} = - \frac{k}{m} x $$

and the differential equation of motion for a simple pendulum is given by the angular acceleration

$$ \frac{d^2\theta}{dt^2} = - \frac{g}{L} \theta $$

where $x$ and $\theta$ are the linear and angular position, respectively, $k$ is the spring constant and $L$ the length of the pendulum. The two equations become equivalent if

$$ \frac{k}{m} \equiv \frac{g}{L}. $$

Putting $x=A\cos(\omega t+\phi)$ in the first equation yields the angular frequency of the simple harmonic motion which then gives us the period of oscillation:

$$ \omega = \sqrt{\frac{k}{m}} \Rightarrow T = 2\pi \sqrt{\frac{m}{k}}. $$

Finally we can establish a relationship between the spring constant and the period:

$$ k_a = 4\pi^2 \frac{m}{T^2}. $$

Here we have calculated the spring constant $k_a$ using the angular motion of the pendulum. But we can calculate the spring constant $k_l$ using the linear simpler harmonic motion of a spring as well. Go back to the equivalence $k/m\equiv g/L$ and imagine now that the length $L$ is not constant but a variable $l$. Then

$$ k_l = \frac{mg}{l}. $$

===== - Experimental method =====
The following apparatus are needed for this experiment.
  - A spring pendulum
  - A stop watch
  - Different masses (50, 100, 150, 200, 250 and 300 gm)

When you attach a mass at the lower end of a vertical spring, the spring will extend and you will calculate the extension of the spring from its equilibrium position for different masses.

Then you will deflect the mass by pulling it to the left side. When you let the mass go, the spring pendulum will go through a simple harmonic motion and you will calculate the time it takes for 10 oscillations of this pendulum. Repeat this step again by deflecting the mass to the right side now. For each repetition calculate the period by dividing the total time by 10.

Once you have the extension $l$ and period $T$ for each mass $m$, you can calculate two different values of the spring constant and the two values should be equal.
===== - Collected data =====

Pendulum period as a function of deflection for $m=150$ gm:
^ Deflection (cm) ^ Time for 10 oscillations $10T$ (s) ^ Mean time $10T$ (s) ^ Period $T$ (s) ^
| 1 |  |  |  | 
| ::: |  | ::: | ::: | 
| 2 |  |  |  | 
| ::: |  | ::: | ::: | 
| 3 |  |  |  | 
| ::: |  | ::: | ::: | 

Pendulum period as a function of mass for a deflection of 2 cm:
^ Mass (gm) ^ Extension $l$ (cm) ^ Time for 10 oscillations $10T$ (s) ^ Mean time $10T$ (s) ^ Period $T$ (s) ^ $T^2$ ^
| 100 |  |  |  |  |  |
| ::: | ::: |  | ::: | ::: | ::: |
| 150 |  |  |  |  |  |
| ::: | ::: |  | ::: | ::: | ::: |
| 200 |  |  |  |  |  |
| ::: | ::: |  | ::: | ::: | ::: |
| 250 |  |  |  |  |  |
| ::: | ::: |  | ::: | ::: | ::: |
| 300 |  |  |  |  |  |
| ::: | ::: |  | ::: | ::: | ::: |

===== - Results and analysis =====

==== Spring constant from angular motion ====

{{ :courses:phy101l:4.1.png?nolink |}}

Period $T$ does not change with changing deflections.

But $T^2$ varies with mass $m$. If $k=4\pi^2m/T^2$ then

$$T^2 = \frac{4\pi^2}{k} m \Rightarrow y = ax + b $$

where $y=T^2$, $x=m$, $b=0$ and the slope of the $T^2-m$ line $a=4\pi^2/k$. This is called the **linearization** of an equation because here we have converted a quadratic equation to a linear equation. If the slope $a$ is known, we can easily calculate the spring constant

$$ k_a = \frac{4\pi^2}{a}. $$

==== Spring constant from linear motion ====
From the equation of the spring constant from linear motion we know

$$ k_l = \frac{mg}{l} \Rightarrow l = \frac{g}{k_l} m \Rightarrow y = ax + b $$

where $y=l$, $x=m$, $b=0$ and the slope of the $l-m$ line $a=g/k_l$. If we know this slope, the spring constant

$$ k_l = \frac{g}{a}. $$


===== - Discussion =====
Compare $k_a$ and $k_l$ and discuss why they should be equal.
===== - Conclusion =====