Car's vibration after being compressed, and released on springs

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SUMMARY

The discussion focuses on calculating the frequency of a car's vibration after it is compressed by the weight of seven individuals averaging 83 kg each. The car's empty mass is 1100 kg, and the compression of the springs is measured at 0.87 cm. The spring constant (k) is calculated to be approximately 1332574.713 N/m, leading to an initial frequency (f1) of 5.34 Hz. However, a miscalculation occurs when attempting to derive the new frequency (f2) for the car's mass using the formula m1f1^2 = m2f2^2, resulting in an incorrect value of 5.539 Hz.

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  • Understanding of Hooke's Law and spring constants
  • Basic knowledge of harmonic motion and frequency calculations
  • Familiarity with gravitational acceleration (9.8 m/s²)
  • Ability to manipulate algebraic equations for physics problems
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  • Study the derivation and application of Hooke's Law in mechanical systems
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rinarez7
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1. When seven people whose average mass is 83 kg sit down in a car they find that the car drops down 0.87cm lower on its springs. Then they get out of the car and bounce it up and down. The acceleration of gravity os 9.8m/s^2.
What is the frequency of the car's vibration if its mass (empty) is 1100kg? Answer in units of Hz.



2. f= 1/2pi (sqrt k/m)

k=F/x= mg/x

m1f1^2 = m2f2^2




3. First I solved for k: k= mg/ x = 1183 kg (9.8) / .0087m= 1332574.713 N/m

Then I solved for f1= 1/2pi (sqrt k/m)= 1/2pi (sqrt (1332574.713/ 1183kg)= 5.34 Hz

Now, when I tried to use these numbers in m1f1^2= m2f2^2 and solve for f2
1183kg(5.34)^2 = 1100kg (f2)^2
f2= 5.539 Hz
But this is wrong. Help!
 
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Hi,

F=kx > G=kx > (83*7)g = k *.0087

f1= 1/2pi (sqrt k/m)= 1/2pi(sqrt(k/1100))
 
Thank you! I missed the 83 kg * 7 people.
 

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