When a 83.0 kg. person climbs into an 1600 kg. car, the car's springs compress vertically 1.6 cm. What will be the frequency of vibration when the car hits a bump? (Ignore damping.)

Answer: the correct answer is f = 0.8747 oscillations/sec

Explanation:

The first bit of info allows you to find the spring constant;

K = force of compression/compression

where the force of compression is the "added" weight on the spring;

K = (83) (9.8) / (.016) K = 50837.5

Once you have, K, you can use the frequency relation between, K, and total mass on spring, M = 1600Kg + 83Kg M=1683 kg, as;

f = (1/2Pi) SqRt[K/M] oscillations/sec

f = (1/2 (3.1416) SqRt (50837.5/1683)

f=1/6.2832) SqRt 30.2065

f = 0.1592 * 5.4960 f = 0.8747 oscillations/sec