Question

A particle moves in xy-plane such that its position as a function of time is given by x = 2sin 2nt, y = 2 (1 - cos 21t) where all parameters are in S.l. units. Then choose the correct statement out of the following Path of the particle is symmetrical about x-axis Net force on the particle is always directed towards origin Speed of the particle is changing with time Average speed of the particle in the time-interval t = 0 to t = 2 sec is 4 mm/sec
​

Answer 1

Answer:

Comparing r = (2sin3t)

i

^

+ 2(1 - cos3t)

j

^

with

r

ˉ

= x

i

^

+ y

j

^

, we have x= 2 sin 3t and y = 2(1 - cos 3t).

This gives sin3t =

2

x

and cos 3t = 1 -

2

y

.

Eliminating t by squaring and adding the above terms, we have

4

x

2

+ (1−

2

y

2

) = 1

Answer 2

Given the position of a particle in xy -plane determine which of the following statements are true about it.

Explanation:

• here we have the position of a particle as a function of time then let the position vector be given as,                                    $\hat r= x \hat i+y \hat j \\x=2sin2t\ \ \ \ \ y=2(1-cos2t)$  
• from above values of x and y we get the trajectory equation of the particle as,                                                                                                            $x=2sin2t\ \ \ \ \ \ \ \ \ \ \ \ \ \ ->sin2t=\frac{x}{2} \\y=2(1-cos2t)\ \ \ \ \ \ \ ->cos2t=1-\frac{y}{2} \\=>sin^22t+cos^22t=1\\=>(\frac{x}{2} )^2+(1-\frac{y}{2})^2=1 \\=>x^2+y^2-4y=0$-----(a)  
• from (a) we get that the path of the particle is not symmetric about x-axis and the net force is not always directed towards origin.
• now the velocity of the particle is given by,                                                          $v_x=\frac{dx}{dt}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_y=\frac{dy}{dt} \\=>v_x=\frac{d(2sin2t)}{dt}\ \ \ \ \ \ \ \ =>v_y=\frac{d(2-2cos2t)}{dt}\\=>v_x=4cos2t\ \ \ \ \ \ \ \ \ \ =>v_y=4sin2t$  
• now the resultant velocity of the particle is given by,                                  $v=\sqrt{v_x^2+v_y^2} \\v=\sqrt{4^2(sin^22t+cos^22t)} \\v=4$
• hence the resultant velocity of the particle is independent of time. Therefor the average speed between $t=0\ to\ t=2$ is ,                                  $s=\frac{v_0+v_2}{2} =\frac{4+4}{2} \\s=4\ m/s$   ->hence third statement is correct.