00:01
Hello students, so here in the question it is given that t is equal to 90 mary second or we can write this as 90 multiplied by 10 square minus 3 second and here is equal to 30 meter per second and this is here along negative x direction right so let's come into the a part of this question.
00:30
So at x is equal to 0.
00:33
Transverse position of the particle, that is y is equal to 2 cm and we can write this as 0 .02 meter.
00:43
And transverse velocity.
00:46
So this is here minus 5 meter per second.
00:53
Now, we know the formula that v is equal to moment.
01:00
Root over a b squared minus y square so this is square both the side so this is omega square a square minus y square so b here this is minus 5 square is equal to omega so this is 4 pi square over t square that is here we have to find and y is 0 .02 square so finally we are getting here that a square minus 4 multiplied by 10x to power minus 4 so this is coming up to the 2 .29 multiplied by 10 x .4 minus 4 as we are getting the value of a amplitude is equal to 2 .5 1 multiplied by 10 x2 minus 2 so this is the answer of a part here now coming to the b part so in the b part here y is equal to a sine omega t plus of five so here we'll put the values y that is equal to 0 .0 and this is equal to 2 .5 1 multiplied by 10 to power minus 2 and here sine of omega t plus 5 t is 0 plus of so from here we are getting the value of 5 is equal to here.
02:47
Or basically we can write sine 5 is equal to 0 .797.
02:54
So from here we are getting 5 is equal to sine inverse of 0 .797 and we are getting 5 is equal to 0 .922 radiance.
03:11
Now, coming to the third part, that is the c part.
03:16
So we have to find the maximum transverse speed.
03:20
So v max is equal to a omega...