A car of mass 1300 kg is constructed using a frame supported by four springs. Each spring has a spring constant 20,000 N m⁻¹. If two people riding in the car have a combined mass of 160 kg, find the frequency of vibration of the car, when it is driven over a pot hole in the road. Assume the weight is evenly distributed. (Ans: 1.18 Hz  0.6 Hz)



Data Given:

Mass of car = m₁ = 1300 kg

Combine Mass of two person = m₂ = 160 kg

Total mass  = m = m₁ + m₂ = 1300 kg + 160 kg = 1460 kg

Spring constant = k = 20,000 N m⁻¹

To Find:


Frequency of the simple pendulum  = f = ?


Solution:

The general formula for Time Period of  mass  attached to spring is

T = 2π `\sqrt frac{m}{K}`

but we know the frequency and time period relationship is

f = `\frac {1}{T}`

thus general formula for frequency for mass attached to spring will be

f = `\frac {1}{2π}` `\sqrt frac{K}{m}`

by putting values 

f = `\frac {1}{2 ｘ 3.1416}` `\sqrt frac{20,000 N m⁻¹}{1460 m}`

f = `\frac {1}{6.292}` ｘ `sqrt 13.699 s⁻²`

f = 0.159ｘ 3.7 s⁻¹

f = 0.5883 Hz   {s⁻¹ = Hz}

or

 f = 0.6 Hz  ------------------Ans.

the answer written in the book is wrong Please correct it.



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