Simple Harmonic Motion (Person gets into car)

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SUMMARY

The discussion centers on calculating the frequency of vertical oscillation for a car and an 85.0 kg passenger, where the car has a mass of 2400 kg and sinks 2.35 cm on its springs. The spring constant (k) was derived as 35483 N/m using the equation mg = kx. The natural frequency of oscillation was calculated using the formula ω = √(k/m), resulting in an angular frequency of 3.78 Hz. However, it was clarified that this value represents angular frequency (ω) rather than the actual frequency (f), which is related by the equation f = ω/(2π).

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ThereIam
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An 85.0 kg person steps into a car of mass 2400 kg, causing it to sink 2.35 cm on its springs. If started into vertical oscillation, and assuming no damping, at what frequency will the car and passenger vibrate on these springs?



w= Sqrt (k/m), mg=kx, perhaps?



The way I treated the problem is with the assumption that the "If started into vertical oscillation" implied that the person/car began to oscillate from the new equilibrium position 2.35 cm below the old one. I then got the spring constant k by saying that 85*9.81 = k0.0235. Then I said the natural frequency of oscillation would be w= Sqrt(k/m) plugging in k = 35483 N/m and m = 2485 kg. Then answer I then got was 3.78 Hz. A problem I just did after this one suggests to me my thinking was wrong. Help please!
 
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Hi ThereTam, welcome to PF.
The answer you got is ω, not the frequency.
What is relation between ω and frequency?
 

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