Question

Stone having mass 0.5 kg rotates in horizontal. It is hanged on 1 m rope. If the tension in the rope is 80 N. Find the frequency of the motion.​

Answer 1

Answer:

frequency (f) = 2 s^-1

Explanation :-

F(net) = 80 N

mass (m) = 0.5 kg

r = 1 m

___________________[Given]

Solution :-

We have,

w – omega

F(net) = mw²r = (m)(4π²f²)(r)

=> 80 = 0.5 × 4 × (3.14)² × f² × 1

=> 80 = 2 × 9.85 × f²

=> f² = 80/19.7

=> f² = 4.06 ≈ 4

=> f = 2 s^-1

____________________________

frequency = 2 sec^-1

____________________[Answer]

Answer 2

hey mate...

here is your answer...

Fnet=80N=m.ω2.r

80=m.4.π2.f2.r

80=0,5.4.32.f2.1

f=2s-1

______________________________________________________

F(net) = 80 N

mass (m) = 0.5 kg

r = 1 m

Solution :-

We have,

w – omega

F(net) = mw²r = (m)(4π²f²)(r)

=> 80 = 0.5 × 4 × (3.14)² × f² × 1

=> 80 = 2 × 9.85 × f²

=> f² = 80/19.7

=> f² = 4.06 ≈ 4

=> f = 2 s^-1

frequency = 2 sec^-1

hope it helps...