PHYSICS HIGH SCHOOL

A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, with the free end of the spring at the pivot. What rotation frequency, in rpm, will cause the spring’s length to stretch by 15%?

Answers

Answer 1
Answer:

Answer:

66 rpm

Explanation:

The period of oscillation is given by

where  T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.

Also, net force is given by

Net force=

where is the elongation, L is original length, is the angular velocity

Substituting the equation of into the above we obtain


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Answers

Answer:

Using Newtons Second law of motion

Explanation:

Newtons Second law of motion states that the rate of change momentum of a system is directly proportional to unbalanced External force acting on it.

if we consider all colliding particles as a system we can see there is no external forces acting on the system. the forces that present here due to collision are internal forces

so As per newtons second law moment should be conserved for the colliding particles.

   

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