Simple Harmonic Motion
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Question : 6 of 23
Marks:
+1,
-0
A particle of mass 1 kg is subjected to a force which depends on the position as
= − k ( x
+ y
) kg ms − 2 with k = 1 kg s − 2 . At time t = 0 , the particle's position
= (
+ √ 2
) m and its velocity
= ( − √ 2
+ √ 2
+
) m s − 1 . Let v x and v y denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z = 0.5 m , the value of ( x v y − y v x ) is m 2 s − 1
[JEE Adv 2022 P2]
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