Question

jee question ........ ​

Answer 1

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Answer:-

Here m=8kg

m=mass,a=amplitude

a=60cm =0.60m

F=120N ,Y=dispalcement=0.30m

k=spring constant

sinceF=KY

K=F/Y=120/0.30=400 N/m

Anguler velocity=w×under root K/m

=under root 400/8

=root 50

=7.07

Time period,T=2pi/w

=2×22/7×7.07

=44.44Sec.

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Answer 2

Answer:

1.256 seconds

is the answer