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Question: Two masses m<sub>1</sub> and m<sub>2</sub> (m<sub>1</sub>\> m<sub>2</sub>) are suspended by two spri...

Two masses m1 and m2 (m1> m2) are suspended by two springs vertically and in equilibrium position, extensions in the springs were same. Both the masses are then displaced in vertical directions by same distances and released. In subsequent motion T1, T2 are their time periods and E1, E2 are energies of oscillations respectively then:

A

(a) T1 = T2; E2 < E

A

(b) T1 > T2; E1 > E2

A

(c) T1 < T2; E1 > E2

A

(d) T1 = T2; E1 > E2

Explanation

Solution

(d)

K1 = m1 g/l; k2=m2 g/l from the equilibrium condition.

Using T = 2π mk\sqrt { \frac { \mathrm { m } } { \mathrm { k } } } T1 = 2π

T2 = 2π

⇒ T1 = T2

E = A2

Since ω and A are same for both and m1 > m2

⇒ E1 > E2.