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Two rigid bodies A and B of masses 1 kg and 2 kg respectively are rigidly connected to a spring of force constant 400 `Nm^(-1)` . The body B rests on a horizontal table. From the rest position , the body A is compressed by 2 cm and then released . Deduce (i) teh frequency of oscillation, (ii) total oscillation energy, (iii) the amplitude of the harmonic vibration of the reaction of the table on body B. |
Answer» Since it is body A which is oscillation , m=1 kg (i) Frequency of osillation , `v=(1)/(2pi)sqrt((k)/(m))` or `v=(1)/(2pi)sqrt((400)/(1)) Hz=(10)/(pi)Hz=3.182 Hz` (ii) Total oscillation energy, `E=(1)/(2)mA^(2)omega^(2)` or `E=(1)/(2)mA^(2)((k)/(m))=(1)/(2)kA^(2)` `=(1)/(2)xx400xx(0.02)^(2) J=0.08 J (because a=2 cm=0.02 m)` Total force acting on the table `=(1+2 ) kg wt=3 kg wt=3xx9.8 N=29.4 N` This is the mean upward reaction. Now, due to oscillation, the maximum tension developed in the spring is given by `F=kA=400xx 0.02 N=8 N` Thus , the amplitude of the vibration of reaction of the table will vary from (29.4+8) N to (29.4)N. |
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