Question

A body is executing SHM. When the displacements from the mean position is 4cm and 5 cm,the corresponding velocities of the body is 10 cm/s and 8 cm/s.Then, the
time period of the body is


(a) 2 π sec
(b) π/2 sec
(c) π sec
(d)3 π/2 sec
​

Answer 1

Q]___?

=> c) π sec

Solution :

Velocity of the particle executing SHM at any ins_tant is defined as the time rate of change of its displacement at that inst_ant. It is given by

v = ω √(a^2 - x^2)

where,

x = displacement of the particle

a = acceleration

ω = angular frequency

CASE I

10 = ω √a^2 - 16 _[i]

CASE II

8 = ω √a^2 - 25 _[ii]

Dividing Eq [ii] by Eq [i] we get

5/4 = (√a^2 - 16 ) / ( a^2 - 25 ) or

25/16 = ( a^2 -16 ) / ( a^2 -25 )

16a^2 - 256 = 25a^2 - 625

a^2 = 369 / 9

Putting the value of a^2 in Eq [i] we get,

10 = ω √ [ (369/9) - 16 ]

ω = (10 × 3) / 15 = 2 rad / s

∴ Time period T= 2π /ω

= 2π/2 = π sec !

Answer 2

Explanation:

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