ncert qn from motion in one dimension

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.

8 Answers

33
Abhishek Priyam ·

Ok 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]

1
varun ·

I 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

62
Lokesh Verma ·

62
Lokesh Verma ·

position 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...

62
Lokesh Verma ·

2nd where the slope decreases or increases..

62
Lokesh Verma ·

this 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

1
jaswinder_1948 ·

please tell me nishant sir whether my solution to trigonometry problem is ok or not

62
Lokesh Verma ·

is there any mistake :O

You should be confident dude :)

Your Answer

Close [X]