Question

pendulum
formulas: period = 1/frequency; frequency = 1/period
period = 2π x √(l/9.8)
pi=3.14
solve for the period if

1. l = 3.5m

1. l= 0.75m

1. define period (related to harmonic motion)

waves
formulas: velocity = wavelength x frequency
label this wave below with: \amplitude\, \wavelength\, \equilibrium\, \crest\, \trough\

Explanation:

Problem 6: Solve for the period when \( L = 3.5 \, \text{m} \)

Step 1: Identify the formula for period

The formula for the period \( T \) of a pendulum is given by \( T = 2\pi \times \sqrt{\frac{L}{9.8}} \), where \( \pi = 3.14 \) and \( L \) is the length of the pendulum.

Step 2: Substitute \( L = 3.5 \, \text{m} \) into the formula

First, calculate \( \frac{L}{9.8} \):

$$ \frac{3.5}{9.8} \approx 0.3571 $$

Then, take the square root of this value:

$$ \sqrt{0.3571} \approx 0.5976 $$

Next, multiply by \( 2\pi \) (where \( \pi = 3.14 \)):

$$ 2 \times 3.14 \times 0.5976 \approx 6.28 \times 0.5976 \approx 3.75 \, \text{seconds} $$

Step 1: Identify the formula for period

The formula for the period \( T \) of a pendulum is \( T = 2\pi \times \sqrt{\frac{L}{9.8}} \), with \( \pi = 3.14 \).

Step 2: Substitute \( L = 0.75 \, \text{m} \) into the formula

First, calculate \( \frac{L}{9.8} \):

$$ \frac{0.75}{9.8} \approx 0.0765 $$

Then, take the square root of this value:

$$ \sqrt{0.0765} \approx 0.2766 $$

Next, multiply by \( 2\pi \) (where \( \pi = 3.14 \)):

$$ 2 \times 3.14 \times 0.2766 \approx 6.28 \times 0.2766 \approx 1.73 \, \text{seconds} $$

In harmonic motion (such as the motion of a pendulum or a mass-spring system), the period is the time taken for one complete cycle of the motion. For a pendulum, it is the time it takes to swing from one side to the other and back to the starting point. In general, for any oscillatory or periodic motion, the period is the duration of one full repetition of the motion.

Answer:

The period of the pendulum when \( L = 3.5 \, \text{m} \) is approximately \( \boldsymbol{3.75 \, \text{s}} \) (or more precisely, following the calculations: \( 2 \times 3.14 \times \sqrt{\frac{3.5}{9.8}} \approx 3.75 \, \text{s} \)).

Problem 7: Solve for the period when \( L = 0.75 \, \text{m} \)