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प्रश्न
Choose the correct option:
A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t(second) is given by x = 6 sin `(100t + π/4)`. Maximum kinetic energy of the body is ______.
पर्याय
36J
9J
27J
18J
उत्तर
A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t(second) is given by x = 6 sin `(100t + π/4)`. Maximum kinetic energy of the body is 18J.
Explanation:
Express the relation for the displacement of a body executing S.H.M.
`x = A sin(omega t + phi)` ......(1)
Here, x is the displacement, `omega` is the angular speed, t is the time in seconds, `phi` is the initial phase of the particle executing S.H.M.
The displacement of a body x(cm) at t(second) is given by,
`x = 6sin(100t + pi/4)` ....(2)
Comparing equation (2) with (1),
`omega = 100`rads–1, `phi = pi/4` and A = `(6 cm xx (10^-2 m)/(1 cm))`
A = `6 xx 10^-2` m
Express the relation for the velocity of a body executing S.H.M.
`v = omegasqrt(A^2 - x^2)` .....(3)
Here, v is the velocity of the body.
Express the relation for the kinetic energy of a body executing S.H.M.
`E_k = 1/2 mv^2` .....(4)
Here, m is the mass of the body.
Substitute value of v from equation (3) in (4),
`E_k = 1/2 m(omegasqrt(A^2 - x^2))^2`
= `1/2 momega^2 (A^2 - x^2)` .....(5)
The kinetic energy is maximum at the mean position, x = 0.
Substitute x = 0, m = 1 kg, `omega = 100`rads–1, and A = `6 xx 10^-2` m in equation (5).
`E_k = 1/2 momega^2 A^2`
= `1/2 xx 1 kg xx ((100"rad")/s)^2 x (6 xx 10^-2 m)^2`
= 18J
Hence, the maximum kinetic energy of the body is 18J.
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