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The new frequency of oscillation when the car bounces on its springs is 0.447 Hz
Frequency of oscillation of spring
The frequency of oscillation of the spring is given by f = (1/2π)√(k/m) where
- k = spring constant and
- m = mass on spring
Now since k is constant, and f ∝ 1/√m.
So, we have f₂/f₁ = √(m₁/m₂) where
- f₁ = initial frequency of spring = 1.0 Hz,
- m₁ = mass of driver,
- f₂ = final frequency of spring and
- m₂ = mass on spring when driver is joined by 4 friends = 5m₁
So, making f₂ subject of the formula, we have
f₂ = [√(m₁/m₂)]f₁
Substituting the values of the variables into the equation, we have
f₂ = [√(m₁/m₂)]f₁
f₂ = [√(m₁/5m₁)]1.0 Hz
f₂ = [√(1/5)]1.0 Hz
f₂ = 1.0 Hz/√5
f₂ = 1.0 Hz/2.236
f₂ = 0.447 Hz
So, the new frequency of oscillation when the car bounces on its springs is 0.447 Hz
Learn more about frequency of oscillation of spring here:
https://brainly.com/question/15318845
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