Question

1. A particle is revolving in U.C.M of radius 4 cm with angular velocity of 10 radian/s. The foot of the perpendicular moves along vertical diameter in S.H.M. The maximum velocity of the foot of the perpendicular is​

Answer 1

Answer:

16

1

s

2

Explanation:

Displacement=x=OQ=Rcosθ

Acceleration=

dt

2

d

2

x

=

dt

2

d

2

(Rcosθ)

=−w

2

Rcosθ

Acceleration

Displacement

=

ω

2

Rcosθ

Rcosθ

=

w

2

1

v=ωR⇒ω=

R

v

=

1 m

4 m/s

=4 rad/sec

Acceleration

Displacement

=

16

1

s

2