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प्रश्न
let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
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उत्तर १
(a) x = 2sin 20t
(b) x = 2cos 20t
(c) x = –2cos 20t
The functions have the same frequency and amplitude, but different initial phases.
Distance travelled by the mass sideways, A = 2.0 cm
Force constant of the spring, k = 1200 N m–1
Mass, m = 3 kg
Angular frequency of oscillation:
`omega = sqrt(k/m)`
`= sqrt(1200/3)= sqrt400 = 20 rad s^(-1)`
a) When the mass is at the mean position, initial phase is 0.
Displacement, x = Asin ωt
= 2sin 20t
b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is `pi/2`
Displacement , `x = Asin(omegat + pi/2)`
`= 2sin (20t + pi/2)`
= 2cos 20t
(c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is `(3pi)/2`
Displacement, `x = Asin(omegat + 3pi/2)`
`= 2sin (20t + 3pi/2) = - 2 cos 20 t`
The functions have the same frequency (`20/(2pi) Hz`) and amplitude (2 cm), but different initial phases `(0, pi/2, (3pi)/2)`
उत्तर २
a =2 cm, omega = `sqrt(k/m) = sqrt(1200/3) s^(-1)= 20s^(-1)`
a) Since time s measured from mean position
b) At the maximum stretched position, tyhe body is at the extreme right position. The initial phase is `pi/2`
`:. x = a sin (omegat + pi/2) = a cos omegat = 2 cos 20 t`
c) At the maximum compressed position, the body is at the extreme left position. The initial phase is `(3pi)/2`
`:. x = a sin (omegat + (3pi)/2) = - a cosomegat = - 2 cos 20t`
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