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Show that the average speed over one oscillation of a particle performing shm of amplitude `a` and angular speed is 2÷3.14 vmax

Sunil sakharam arsul , 9 Years ago
Grade 12
anser 2 Answers
Saloni manoj bhosale
Given,the particle performing shm in one oscillation of amplitude 'a'.
The particle covers distance of 4a in one oscillation.
V=4a/T
But,T=2×3.14/W
V=4aW/2×3.14
V=2/3.14×aW
V=2/3.14Vmax
 
Last Activity: 6 Years ago
Vikas TU
Dear Student,
Please find below the solution to the asked query:
The distance travelled in one complete oscillation by a particle performing SHM is 4A,
Then, the average velocity over a time period is,
Vavg= Total distance /time 
4a / (2*pie/w) = 4aw/2*pie 
Vavg = 2aw/pie
Last Activity: 6 Years ago
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