a) An 86 kg person steps into a car of mass 2437 kg, causing it to sink 2.35 cm

Question

a) An 86 kg person steps into a car of mass 2437 kg, causing it to sink 2.35 cm on its springs. Assuming no damping, with what frequency will the car and passenger vibrate on the springs?

b) A 0.117-kg block is suspended from a spring. When a small pebble of mass 30 g is placed on the block, the spring stretches an additional 5.1 cm. With the pebble on the block, the block oscillates with an amplitude of 12 cm. Find the maximum amplitude of oscillation at which the pebble will remain in contact with the block.

Explanation / Answer

a)

Weight of the car is balanced by the spring force:

So, mg = kx

where x = initial compression of the pring when the person is not there .

So, k = mg/x = 2437*9.8/x

when the person comes, x' = x + 0.0235

and m = 2437+86

So, (2437+86)*9.8/(x+0.0235) = 2437*9.8/x

So, x = 0.6659 m

So, k = 2437*9.8/0.6659 = 3.587*10^4 N/m

So, frequency of oscillation, f = (1/(2*pi))*sqrt(k/m)

So, f = (1/(2*pi))*sqrt(3.587*10^4/(2437+86)) = 0.6 Hz

b)

F = k*x = Mg

where x = initial extension

M = 0.117 kg

when m = 30 g(=0.030 kg) is added,

k*(x+0.051) = (0.117+0.03)*9.8

So, x/(x+0.051) = (0.117)/(0.117+0.03)

So, x = 0.199 m = 19.9 cm

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