Question

6. A particle executing SHM oscilates between two fixed points separated by 20 cm
If its maximum velocity is 30 cm/s. then find its velocity when its displacement is 5 cm
from its mean position.
d. 15√ 3 m/s
b. 15 m/s
c. √2 m/s
d. 10 m/s​

Answer 1

Answer:

• this can be solved by using phaser diagram . amplitude is 10cm. maximum velocity of particle is AW. AW=30 so W=3 . velocity at any instant is given by formula v = w× under root A square- x square
• this can be solved by using phaser diagram . amplitude is 10cm. maximum velocity of particle is AW. AW=30 so W=3 . velocity at any instant is given by formula v = w× under root A square- x square v= 3 under root 100- 25
• this can be solved by using phaser diagram . amplitude is 10cm. maximum velocity of particle is AW. AW=30 so W=3 . velocity at any instant is given by formula v = w× under root A square- x square v= 3 under root 100- 25 =3 under root 75
• this can be solved by using phaser diagram . amplitude is 10cm. maximum velocity of particle is AW. AW=30 so W=3 . velocity at any instant is given by formula v = w× under root A square- x square v= 3 under root 100- 25 =3 under root 75 = 15 under root 3
• hope it helps you