Question

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 1

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

Step-by-step 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