A `1kg` block is executing simple harmonic motion of amplitude `0.1m` on a smooth horizontal surface under the restoring force of a spring of spring constant `100 N//m`. 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 and the amplitude of the motion.

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