The length of a simple pendulum is 0.79m and the mass of the particle at the end of the cable is 0.24 kg. The pendulum is pulled away from its equilibrium position by an angle of 8.5 degrees and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. What is the angular frequency of the motion? What is bobs speed as it passes through the lowest point of the spring?
STEPS PLEASE. answer doesnt matter.
The angular frequency of the harmonic motion is
w = sqrt (g/L) radians per second,
and is independent of mass. Calculate it.
This is an accurate approximate formula that is almost always used. The approximation
sin theta = theta
is made in the derivation.
There is a small correction for angles greater than about 15 degrees.
The bob's maximum horizontal deflection from vertical is
A = L sin 8.5 = 0.117 m
The bob's x position is
x = A cos wt
The maximum velocity (at the lowest point) is w *A
To find the angular frequency of the motion, we can use the formula:
ω = √(g/L)
where:
- ω is the angular frequency in radians per second,
- g is the acceleration due to gravity (approximately 9.8 m/s²), and
- L is the length of the pendulum.
1. Substitute the given values into the formula:
ω = √(9.8 / 0.79)
2. Calculate the square root and divide:
ω = √(12.405¹²⁵ / 0.623²⁵) ≈ √19.8614 ≈ 4.46 rad/s
The angular frequency of the motion is approximately 4.46 radians per second.
To find Bob's speed as it passes through the lowest point of the pendulum swing, we can use the formula for velocity in simple harmonic motion:
v = ω * A
where:
- v is the velocity,
- ω is the angular frequency, and
- A is the amplitude of the oscillation.
In this case, the amplitude is the distance from the equilibrium position to the maximum displacement, which is equal to the length of the pendulum (0.79 m).
3. Substitute the values into the formula:
v = 4.46 * 0.79
4. Multiply to find the velocity:
v ≈ 3.52 m/s
Bob's speed as it passes through the lowest point of the pendulum swing is approximately 3.52 meters per second.
Step 1: Identify the given parameters:
- Length of the pendulum (L) = 0.79m
- Mass of the particle (m) = 0.24kg
- Angle of displacement (θ) = 8.5 degrees
Step 2: Convert the angle from degrees to radians:
- θ (in radians) = θ (in degrees) × π / 180
- θ (in radians) = 8.5 × π / 180
Step 3: Calculate the gravitational acceleration (g):
- g = 9.8 m/s^2 (approximately)
Step 4: Calculate the period (T) of the pendulum:
- T = 2π √(L / g)
Step 5: Calculate the angular frequency (ω) of the motion:
- ω = 2π / T
Step 6: Calculate the speed (v) of the particle as it passes through the lowest point:
- v = L * ω
Note: The above steps assume that the pendulum is undergoing simple harmonic motion.
Please note that this is the general procedure to solve the given problem. The specific values and calculations may vary based on the actual parameters.