Motion in a Straight Line
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Question : 20
Total: 27
Figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
Solution:
We know that in S.H.M., the acceleration ‘a’ is given by therelation,
a = − ω 2 x . . . .
where w is the angular frequency and is a constant. Also velocity is givenby,
v =
. . . .
(i) At time t = 0.3 s, x is negative, slope of (x-t) plot is negative, soposition and velocity are negative but acceleration is positive accordingto equation (i).
(ii) At time, t = 1.2 s, x is positive, the slope of (x-t) plot is also positive,hence position and velocity are positive but according to equation (i),acceleration is negative.
(iii) At time, t = –1.2s, x is negative, the slope of (x-t) plot is also negative.But since both x and t are negative hence according to equation (ii),velocity is positive. As position is negative, so according to equation (i)acceleration is positive.
where w is the angular frequency and is a constant. Also velocity is givenby,
(i) At time t = 0.3 s, x is negative, slope of (x-t) plot is negative, soposition and velocity are negative but acceleration is positive accordingto equation (i).
(ii) At time, t = 1.2 s, x is positive, the slope of (x-t) plot is also positive,hence position and velocity are positive but according to equation (i),acceleration is negative.
(iii) At time, t = –1.2s, x is negative, the slope of (x-t) plot is also negative.But since both x and t are negative hence according to equation (ii),velocity is positive. As position is negative, so according to equation (i)acceleration is positive.
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