1please tell me nishant sir whether my solution to trigonometry problem is ok or not
3.20 Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple
harmonic motion. (You will learn about this motion in more detail in Chapter14).
Give the signs of position, velocity and acceleration variables of the particle at
t = 0.3 s, 1.2 s, – 1.2 s.
I dont kw how to drw the diagrm here. so pls chek ncert class-xi part- 1, pg-58 , qn-3.20.
I m not able to find 'acceleration' in this qn. pls explain me how to find.
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UP 0 DOWN 0 0 8
8 Answers
33Ok dear i have to first search for class XI book. Let me ....
And drawing is concerned u can draw it in paint or any other prog and insert image here...
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1I think -
until the mean point, first acc increases from one end to mean point and then decreases from there to the other end.
for t=0.3s
It is decreasing but it is in the negative dir... so a >0
for t=1.2s,
it is decreasing as it has passed the mean point.. so a <0
for t=-1.2s
it is before the mean position,i.e. it is getting accelerated .. so a>0
62position will be +ve if x > 0
velocicty is +ve if dx/dt> 0
acceleration is +ve if d2x/dt2 > 0
1st is solved by the points where the blue line is above the x axis...
62this is a sin graph..
so the double derivative will be
-ve sign of f(x)
so "in this case only" the acceleration will be exactly of opposite sign of displacemnet
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