A 1kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring constant `100Nm^-1`. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency.

Answer

B, C

Explanation

Amplitude =0.1m
total mass =3+21=4kg ltbrrgt (when both the blocks are moving together)
`K=100 N/M`
`:. T=2pi m/k`
`=2pisqrt(4/100)=(2pi)/5 sec`
`:. Frequencey =5/(2pi) Hz`
Again at the mean position let 1 kg block has velocity v
`KE=1/2mv^2=1/2kx^2`
Where xrarr Amplitude =0.1m
`:. (1/2)xx(1xxv^2)=(1xx2)xx100(0.1)^2`
`Kv=1msec^-1..........i `
After the 3 kg block is gently place on teh 1 kg, then let 1kg+3kg-4kg block and the spring the be one system. For this mass spring system there is no extgernla force. (when oscillation takes place). The momentum should be conserved. Let 4 kg block has velocity v.
`:. Intial momentum =FiN/Al momentum
`1xxv=4xxv^1`
`rarr v^1=1/4m/s`
`[as v=1 msec^-1
at its mean position
`KE mean =1/2mv^2`
`=(1/2)xx4xx(1/4)^2=1/2xx1/4`
`KE mean `=(1/2)m^1v^2(1/2)4x(1/4)^2`
`=1/2xx1/4`
when the blocks are going to the extreme position, there will be only potential energy.
`:. PE=1/2kdelta^2=1/2xx1/4`
where drarr New amplitude
`:. 1/4=100delta^2`
`=d=sqrt((1/400))`
`=0.05m=5m`
`So, amplitude =5cm