A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is

Options

A. `(1)/(4)Hz`
B. `(1)/(2sqrt(2))Hz`
C. `(1)/(2)Hz`
D. `2Hz`

Answer

C

Explanation

Frequency of spring (f) `=(1)/(2pi)sqrt((k)/(m))=1" Hz"`
`implies" "4pi^(2)=(k)/(m)`
If block of mass m = 1 kg is attached then, `k=4pi^(2)` Now, identical springs are attached in parallel with mass m = 8 kg. Hence, `k_(eq)=2k`
`F=(1)/(2pi)sqrt((kxx2)/(g))=(1)/(2)Hz`