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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.)

Respuesta :

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

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