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A spring has force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is suspended from one of the halves. How are the frequencies of oscillation , before and after the spring is cut, related?

A spring has force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is suspended from one of the halves. How are the frequencies of oscillation , before and after the spring is cut, related?

Grade:12th pass

2 Answers

Himanshu
389 Points
8 years ago
the spring constant is inversely propotional to the lenght....
hence whrn a spring is cut into two equal halves the spring constant of both will become two times....as lenght of spring is halved...time period of osccilations 2(pi)underoot(m/k) hence mass is unchanged putting values of intial and final k ratio of original/new time period is: 1.414:1 or root(2):1
shubham rathour
11 Points
8 years ago
we know spring constant k is proportional to natural length of spring ..so when a spring is cut length becomes half and as a result spring constant becomes double and frequency of oscillation is directly proportional to uder root of spring constant ..as a result new frequency becomes 1.414 times the older..
 

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