A particle of mass m = 2 kg executes SHM in xy-plane between points A and B under action of force = Fx + Fy . Minimum time taken by particle to move from A to B is 1 sec. At t = 0 the particle is at x = 2 and y = 2. Then Fx as function of time t is
ans -4∩2 cos∩t
please explain
1Okie...lemme try.
Let the line joining AB represent an axis 'r'. (+r is in the direction of +ve inclination of x-axis).
By the condition given, r co-ordinate of the particle at time t is
r=2\sqrt{2}cos\omega t
amplitude bieng 2√2
\omega =\frac{2\pi }{T}=\pi
\Rightarrowr=2\sqrt{2}cos\pi t
x= rcos \frac{\pi }{4}=\frac{r}{\sqrt{2}}=2cos\pi t
Therefore,
a_{x}=-\omega ^{2}x
\Rightarrow -2\pi ^{2}cos \pi t
F_{x}=ma_{x}=-4\pi ^{2}cos\pi t
1see the particle oscillating about line ab.
then equation of particle is : 2√2cos ∩t.
and since time period is : 2 sec
hence accleration is : a=-2√2∩2cos∩t
force along this line is : √2F
HENCE EQUATING √2F=m a.WE GET THE DESIRED RESULT
1those seem correct. Thanks to both of u
(with the slight exception of the missing 2 in the expression for acceleration..but I got the point)
[1][1]