Simple Harmonic Motion

Section: Physics
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Question : 6 of 23
 
Marks: +1, -0
A particle of mass 1kg is subjected to a force which depends on the position as
F
=
k(x
ı
+y
)
kg
ms2
with k=1kgs2. At time t=0, the particle's position
r
=
(
1
2
ı
+2
)
m
and its velocity
v
=(2
ı
+2
+
2
π
k
)
m
s1
. Let vx and vy denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z=0.5m, the value of (xvyyvx) is m2s1
[JEE Adv 2022 P2]
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