MHTCET 2016 Physics Solved Paper

(a) For a given oscillating mass, potential energy is given by
PE =

1

2

kx2
For a body oscillating at x = A, maximum energy is given by
Emax=

1

2

KA2
Also at mean position x = 0.
So, E = 0
∴ A ∝

1

√m

....(i)
When another mass m2 is placed on mass m1. Then, total mass becomes (m1+m2) and at this point E = 0 as x = 0.
When they reach at x = A,
A1∝

1

√m1+m2

....(ii)
Dividing eq. (ii) by eq. (i),

A1

A

=[

m1

m1+m2

]

1

2