A weightless spring which has a force constant oscillates with frequency n when a mass m is suspended from
it. The spring is cut into two equal halves and a mass 2m is suspended from it. The frequency of oscillation will
now become
(A)n
(B) 2n
(C) n/2
(D) n(2)2
Answers
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Answer:
D is the answerbgfhshbssj
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Since time period remains same in both cases, frequency is also same.
Correct option is (A).
Explanation:
Time period of oscillation of a body of mass "m" suspended by a spring is
T = 2π √ m / k
On cutting spring in two equal parts, the length of each part will remain half and the force constant will be doubled (2k).
Therefore,
T = 2π √ 2m / 2k = f
Since time period remains same in both cases, frequency is also same.
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