Question

stopwatch and time how long it takes for the spring to move from top to bottom times.
a. record the time in the table below and complete the other calculations.
data and calculations: trial 1

time	period	frequency	wavelength

directions; trial 2

1. keep the slinky stretched to 50cm.
2. increase the speed of the moving end of the slinky until the path of the slinky resembles the image to the right.
3. take a stopwatch and time how long it takes for the spring to move through 10 cycles.
4. record the time in the data table below and complete the other calculations.

data and calculations: trial 2

time	period	frequency	wavelength

Explanation:

Trial 1 Calculations:

Step 1: Calculate Period (\(T\))

Given \( t = 11 \, \text{s} \) (time for 10 cycles), use \( T = \frac{t}{10} \).
\( T = \frac{11}{10} = 1.1 \, \text{s} \)

Step 2: Calculate Frequency (\(f\))

Use \( f = \frac{1}{T} \). Substitute \( T = 1.1 \, \text{s} \).
\( f = \frac{1}{1.1} \approx 0.91 \, \text{Hz} \)

Trial 2 Calculations:

Step 1: Calculate Period (\(T\))

Given \( t = 19 \, \text{s} \) (time for 10 cycles), use \( T = \frac{t}{10} \). Wait, the handwritten \( T = 1.9 \, \text{s} \) implies \( t = 19 \, \text{s} \) (since \( \frac{19}{10} = 1.9 \)).

Step 2: Calculate Frequency (\(f\))

Use \( f = \frac{1}{T} \). Substitute \( T = 1.9 \, \text{s} \).
\( f = \frac{1}{1.9} \approx 0.53 \, \text{Hz} \)

Final Answers (Filled Table):

Trial 1:

• \( t = 11 \, \text{s} \)
• \( T = 1.1 \, \text{s} \)
• \( f \approx 0.91 \, \text{Hz} \)
• \( \lambda = 100 \, \text{cm} \)

Trial 2:

• \( t = 19 \, \text{s} \)
• \( T = 1.9 \, \text{s} \)
• \( f \approx 0.53 \, \text{Hz} \)
• \( \lambda = 50 \, \text{cm} \)

(Note: If the handwritten \( t \) for Trial 2 is different, adjust \( T \) and \( f \) accordingly using \( T = t/10 \) and \( f = 1/T \).)

Answer:

Trial 1 Calculations:

Step 1: Calculate Period (\(T\))

Given \( t = 11 \, \text{s} \) (time for 10 cycles), use \( T = \frac{t}{10} \).
\( T = \frac{11}{10} = 1.1 \, \text{s} \)

Step 2: Calculate Frequency (\(f\))

Use \( f = \frac{1}{T} \). Substitute \( T = 1.1 \, \text{s} \).
\( f = \frac{1}{1.1} \approx 0.91 \, \text{Hz} \)

Trial 2 Calculations:

Step 1: Calculate Period (\(T\))

Given \( t = 19 \, \text{s} \) (time for 10 cycles), use \( T = \frac{t}{10} \). Wait, the handwritten \( T = 1.9 \, \text{s} \) implies \( t = 19 \, \text{s} \) (since \( \frac{19}{10} = 1.9 \)).

Step 2: Calculate Frequency (\(f\))

Use \( f = \frac{1}{T} \). Substitute \( T = 1.9 \, \text{s} \).
\( f = \frac{1}{1.9} \approx 0.53 \, \text{Hz} \)

Final Answers (Filled Table):

Trial 1:

• \( t = 11 \, \text{s} \)
• \( T = 1.1 \, \text{s} \)
• \( f \approx 0.91 \, \text{Hz} \)
• \( \lambda = 100 \, \text{cm} \)

Trial 2:

• \( t = 19 \, \text{s} \)
• \( T = 1.9 \, \text{s} \)
• \( f \approx 0.53 \, \text{Hz} \)
• \( \lambda = 50 \, \text{cm} \)

(Note: If the handwritten \( t \) for Trial 2 is different, adjust \( T \) and \( f \) accordingly using \( T = t/10 \) and \( f = 1/T \).)