Question

maximum velocity of particle in shm is 2 cm/s then magnitude of average velocity during one extreme position to another extreme position is​

Answer 1

Dear Student,

◆ Answer -

vavg = 1.273 m/s

◆ Explanation -

Maximum velocity of a particle in SHM is given by -

vmax = ωA

vmax = 2πA/T

2 = 2πA/T

A/T = 2/2

A/T = 1/π

Average velocity of particle from one extreme position to another is -

vavg = distance / time

vavg = 2A / (T/2)

vavg = 4 A/T

vavg = 4 × 1/π

vavg = 4 / 3.142

vavg = 1.273 m/s

Therefore, average velocity during one extreme position to another extreme position is 1.273 m/s.

Thanks dear. Hope this helps you...

Answer 2

Explanation:

so the magnitude of average velocity during one extreme position to another position is 4/pi.Hope this helps you