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Home » In the given figure, let us take the position of mass when the spring is unstreched as , and the direction from left to right as the positive direction of -axis. Give as a function of time for the oscillating mass if at the moment we start the stopwatch (t , the mass is(a) at the mean position,(b) at the maximum stretched position.
In the given figure, let us take the position of mass when the spring is unstreched as x=0, and the direction from left to right as the positive direction of 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.
CBSE | Class 11 | NCERT | Oscillations | Physics
In the given figure, let us take the position of mass when the spring is unstreched as x=0, and the direction from left to right as the positive direction of 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.

Solution:

Distance travelled by the mass sideways is given as a=2.0 \mathrm{~cm}

Angular frequency of oscillation can be calculated as,

\omega=\sqrt{k} / \mathrm{m}

=\sqrt{1200 / 3}

=\sqrt{400}

=20 \mathrm{rad} \mathrm{s}^{-1}

(a) As time is noted from the mean position,

So, using

x=a \sin \omega t

We have,

x=2 \sin 20 t

(b) At maximum stretched position, the body is at the extreme right position, with an initial phase of \pi /2 rad. Then,

x=a \sin (\omega t+m / 2)

=\mathrm{a} \cos \omega \mathrm{t}

=2 \cos 20 \mathrm{t}

i

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