A boy with a mass of 50 kg is hanging from a spring with
a spring constant of 200 N/m. With what frequency does
the boy bounce up and down?
y = A sin 2 pi f t
y"=-A(2 pi f)^2sin 2pift = -(2 pi f)^2 y
-k y = m y"
- k y = - m (2 pi f)^2 y
(2 pi f)^2 = k/m
f = (1/2pi)sqrt(k/m)
f = (1/2pi) sqrt (200/50)
f = 1/pi = .3183098852 Hz
call it 0.318 Hz
To find the frequency at which the boy bounces up and down, we can use the formula:
f = 1 / (2π) * √(k / m)
where:
- f represents the frequency
- k represents the spring constant
- m represents the mass of the object
Let's substitute the given values into the formula:
f = 1 / (2π) * √(200 N/m / 50 kg)
Now, let's simplify the equation:
f = 1 / (2π) * √(4 N/kg)
f = 1 / (2π) * 2 N/kg
f = 1 / π N/kg
Therefore, the boy bounces up and down with a frequency of approximately 0.318 Hz (Hertz).
To find the frequency at which the boy bounces up and down, you can use the following formula:
f = 1 / (2π) * sqrt(k / m)
where:
- f is the frequency in hertz (Hz)
- k is the spring constant in newtons per meter (N/m)
- m is the mass of the boy in kilograms (kg)
- π is a mathematical constant approximately equal to 3.14159...
Given:
- m = 50 kg
- k = 200 N/m
Substituting these values into the formula, we get:
f = 1 / (2π) * sqrt(200 / 50)
Now, let's calculate:
f = 1 / (2π) * sqrt(4)
Simplifying further:
f = 1 / (2π) * 2
f = 1 / π
Using an approximation of π as 3.14159:
f ≈ 0.3183 Hz
Therefore, the boy bounces up and down with a frequency of approximately 0.3183 Hz.