A tray of mass M=10kg is supported on two identical springs, each of spring constant k, as shown in figure, when the tray is depressed a little and released, it executes simple harmonic motion of period 1.5 s. when a block of mass m is placed on the tray, the speed of oscillation becomes 3 s. the value of m is

Options

A. 10 kg
B. 20 kg
C. 30 kg
D. 40 kg

Answer

C

Explanation

As the two identical springs are connected in parallel,
`therefore`The effective spring constant of the combination is
`k'=k+k=2k`
`thereforeT=2pisqrt((M)/(k'))=2pisqrt((M)/(2k))` . . . (i)
When a block of mass m is placed in the tray, the period of oscillation becomes
`T'=2pisqrt((M+m)/(k'))=2pisqrt((M+m)/(2k))` . . (ii)
Divide (ii) by (i), we get
`(T')/(T)=sqrt((M+m)/(M))`
`(3)/(1.5)=sqrt((10+m)/(10))`
Squaring both sides, we get
`4=(10+m)/(10)` or `40=10+m` or `m=30kg`