Periodic Motion Worksheet - Fill and Sign Printable Template Online - Free Printable

I'll solve each problem on this periodic motion worksheet step by step.

Problem 1: Find the period of a pendulum that is 1.2 m long.

The formula for the period of a pendulum is:
T = 2π√(L/g)

Where:
- T = period (in seconds)
- L = length of pendulum (in meters)
- g = acceleration due to gravity = 9.8 m/s²
- π ≈ 3.14159

Let me calculate:
T = 2π√(1.2/9.8)
T = 2π√(0.1224)
T = 2π × 0.3499
T = 2 × 3.14159 × 0.3499
T = 6.28318 × 0.3499
T ≈ 2.20 seconds

Problem 2: A pendulum has a period of 1.2 s. How long is it?

Using the same formula, I need to solve for L:
T = 2π√(L/g)

Rearranging:
T/(2π) = √(L/g)
(T/(2π))² = L/g
L = g × (T/(2π))²

L = 9.8 × (1.2/(2×3.14159))²
L = 9.8 × (1.2/6.28318)²
L = 9.8 × (0.191)²
L = 9.8 × 0.0365
L ≈ 0.358 m or about 0.36 m

Problem 3: A pendulum has a frequency of 1.2 Hz. Find the period and the length of this pendulum.

First, find the period:
Frequency (f) and period (T) are related by: f = 1/T

So: T = 1/f = 1/1.2 = 0.833 seconds

Now find the length using T = 2π√(L/g):
0.833 = 2π√(L/9.8)
0.833/(2π) = √(L/9.8)
0.833/6.28318 = √(L/9.8)
0.1326 = √(L/9.8)
(0.1326)² = L/9.8
0.0176 = L/9.8
L = 9.8 × 0.0176
L ≈ 0.172 m or about 0.17 m

Problem 4: You are stranded on an island and need to make a clock. How would you build a pendulum that has a period of 1.0 second and another a period of 1.5 minutes?

For T = 1.0 second:
L = g × (T/(2π))²
L = 9.8 × (1.0/6.28318)²
L = 9.8 × (0.159)²
L = 9.8 × 0.0253
L ≈ 0.248 m or about 0.25 m (25 cm)

For T = 1.5 minutes = 90 seconds:
L = 9.8 × (90/6.28318)²
L = 9.8 × (14.32)²
L = 9.8 × 205.1
L ≈ 2010 m or about 2.0 km

(Note: A 1.5-minute pendulum would be extremely long - over 2 kilometers!)

Problem 5: While on the island you discover the tallest tree. To determine its height you suspend a rope from the top and put a weight at the bottom. You pull off and return to the starting point 12 seconds later. How tall is the tree?

The time given (12 seconds) is the period T of the pendulum.

Using T = 2π√(L/g), solve for L:
12 = 2π√(L/9.8)
12/(2π) = √(L/9.8)
12/6.28318 = √(L/9.8)
1.91 = √(L/9.8)
(1.91)² = L/9.8
3.65 = L/9.8
L = 9.8 × 3.65
L ≈ 35.8 m or about 36 m

Problem 6: You discover a spring and want to know its properties. You hang it from a tree and measure the length of the spring to be 1.0 m. You suspend 10 kg from the spring and it stretches to a new length of 1.4 m. Find the spring constant for this spring.

The spring stretched from 1.0 m to 1.4 m, so the displacement x = 1.4 - 1.0 = 0.4 m

Using Hooke's Law: F = kx

Where:
- F = force (weight of the mass)
- k = spring constant
- x = displacement

The force is the weight: F = mg = 10 kg × 9.8 m/s² = 98 N

So: 98 = k × 0.4
k = 98/0.4
k = 245 N/m

Final Answer:
1. Period = 2.2 seconds
2. Length = 0.36 m
3. Period = 0.83 seconds, Length = 0.17 m
4. For 1.0 s period: length = 0.25 m; For 1.5 min period: length = 2010 m (2.0 km)
5. Tree height = 36 m
6. Spring constant = 245 N/m