Oscillations - Result Question 48

52. A body of mass M, executes vertical SHM with periods t1 and t2, when separately attached to spring A and spring B respectively. The period of SHM, when the body executes SHM, as shown in the figure is t0. Then

[2002]

(a) t01=t11+t21

(b) t0=t1+t2

(c) t02=t12+t22

(d) t02=t12+t22

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Answer:

Correct Answer: 52. (d)

Solution:

  1. (d) The time period of spring mass system,

T=2πmK

t1=2πmk1

(ii)t2=2πmk2

When springs are connected in parallel then Keff =k1+k2

So, t0=2πmkeff2πm(k1+k2)

from (i), 1t12=14π2×k1m

from (ii), 1t22=14π2×k2m

from (iii), 1t02=14π2×(k1+k2)m

from above expressions,

1t02=1t12+1t22

t=2πmk

k= Const. t2

Here the springs are joined in parallel. So

k0=k1+k2

where k0 is resultant force constant

Const. t02= Const. t12+ Const. t22

or, t02=t12+t22



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