Question

1.
In SHM velocity of the particle at the mean
position is 2m/s and acceleration at the extreme
position is 4m/s. The angular velocity of the
particle is
1.1 rad s 2.2 rad s 3.4 rad s 4.3 rad s​

Answer 1

Explanation:

We know that for a particle undergoing SHM with an amplitude 'a', velocity 'v' and angular velocity 'ω'

v=ω

a

2

−x

2

At mean position x=a

And given v = 2ms

−1

at mean position

2=ω

a

2

=ωa

v=ω

a

2

−

4

a

2

=ω

4

3a

2

=

2

3

aω

Dividing,

v

2

=

2

ωa

3

ωa

=

3

2

v=

3

ms

−1

=1.732ms

−1

I hope you like it

Answer 2

Answer:

The maximum acceleration for a particle executing SHM is