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...
[1]
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.
-
UP 0 DOWN 0 0 8
8 Answers
331I 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
1please tell me nishant sir whether my solution to trigonometry problem is ok or not