This question outlines a physics experiment involving a mass-spring system and asks for a graph to be plotted based on the results. Since no experimental data is provided, I will explain how to perfor

This question outlines a physics experiment involving a mass-spring system and asks for a graph to be plotted based on the results. Since no experimental data is provided, I will explain how to perform the experiment and what the expected results and graph would look like.

The experiment involves two setups:

• Setup 1 (Fig. 1(a)): A single spring with varying masses.
• Setup 2 (Fig. 1(b)): Two identical springs in parallel with varying masses.

The goal is to measure the period of oscillation ($T$ and $T_0$) and its square ($T^2$ and $T_0^2$) for different masses ($m$).

Part 1: Single Spring System (Fig. 1(a))

• Steps (i) - (vii):
  • Step 1: Suspend a mass $m = 150 \, \text{g}$ (or $0.150 \, \text{kg}$) from a single spring.
  • Step 2: Pull the mass down slightly and release it to set it into vertical oscillation.
  • Step 3: Use a stopwatch to measure the time $t$ for 10 complete oscillations.
  • Step 4: Calculate the period $T$ using the formula $T = \frac{t}{10}$.
  • Step 5: Calculate $T^2$.
  • Step 6: Repeat steps 1-5 for four other masses: $m = 200 \, \text{g}$, $250 \, \text{g}$, $300 \, \text{g}$, and $350 \, \text{g}$.

Part 2: Two Identical Springs in Parallel System (Fig. 1(b))

• Steps (viii) - (xiv):
  • Step 1: Set up the two identical springs in parallel as shown in Fig. 1(b).
  • Step 2: Suspend a mass $m = 150 \, \text{g}$ (or $0.150 \, \text{kg}$) from this spring system.
  • Step 3: Pull the mass down slightly and release it to set it into vertical oscillation.
  • Step 4: Use a stopwatch to measure the time $t_0$ for 10 complete oscillations.
  • Step 5: Calculate the period $T_0$ using the formula $T_0 = \frac{t_0}{10}$.
  • Step 6: Calculate $T_0^2$.
  • Step 7: Repeat steps 1-6 for four other masses: $m = 200 \, \text{g}$, $250 \, \text{g}$, $300 \, \text{g}$, and $350 \, \text{g}$.

Step (xv): Tabulate the results.

A table should be created to record the measurements and calculated values. It should include columns for mass ($m$), time for 10 oscillations ($t$ or $t_0$), period ($T$ or $T_0$), and the square of the period ($T^2$ or $T_0^2$).

Example Table Structure:

Mass $m$ (g)	Time $t$ (s) (Single Spring)	Period $T$ (s) (Single Spring)	$T^2$ (s$^2$) (Single Spring)	Time $t_0$ (s) (Parallel Springs)	Period $T_0$ (s) (Parallel Springs)	$T_0^2$ (s$^2$) (Parallel Springs)
150
200
250
300
350

Step (xvi): Plot a graph of $m$ on the vertical axis and $T^2$ on the horizontal axis.

The relationship between the period of oscillation ($T$) and mass ($m$) for a mass-spring system is given by:

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

where $k$ is the spring constant.

Squaring both sides, we get:

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

Rearranging to match the requested graph axes ($m$ on vertical, $T^2$ on horizontal):

$$m = \left(\frac{k}{4\pi^2}\right) T^2$$

This equation is in the form $y = cx$, where $y = m$, $x = T^2$, and the slope $c = \frac{k}{4\pi^2}$.

Therefore, when you plot $m$ (in kg) on the vertical axis against $T^2$ (in s$^2$) on the horizontal axis, you should obtain a straight line passing through the origin.

• Graphing Instructions:

  • Draw two perpendicular axes.
  • Label the vertical axis as "$m$ (kg)" and the horizontal axis as "$T^2$ (s$^2$)".
  • Choose appropriate scales for both axes to accommodate all your data points.
  • Plot the data points for both the single spring system ($m$ vs. $T^2$) and the parallel spring system ($m$ vs. $T_0^2$). You might use different symbols or colors for each set of data.
  • Draw the best-fit straight line through the origin for each set of data points.

• Expected Outcome:

  • For both systems, the graph of $m$ versus $T^2$ should be a straight line passing through the origin.
  • The slope of the line for the single spring system will be $\frac{k_1}{4\pi^2}$, where $k_1$ is the spring constant of the single spring.
  • For two identical springs in parallel, the effective spring constant $k_{eff} = k_1 + k_2$. Since the springs are identical, $k_{eff} = 2k_1$.
  • Therefore, the slope of the line for the parallel spring system will be $\frac{2k_1}{4\pi^2}$, which is twice the slope of the single spring system. This means the line for the parallel springs will be steeper.

This concludes the explanation of how to perform the experiment and plot the graph.