Question

If the mass of the object were doubled (keeping a constant radius on the string), what would have to happen to the period in order to maintain the same tension on the string

Answer 1

Answer:

the period of the motion will increase by√2.

Explanation:

Given that the motion you are talking about is circular motion of a mass attached to the end of a string. I speculated that from the word usage in the question(e.g radius instead of length, tension, period). Given this is so we will have to recall the formula for the centripetral force Fc acting on the object which will be equal in magnitude to the tension in the string and will be given by,

$F_{c} = mrw^{2}$

if we want the above defined tension to remain constant when we double the mass and keep the radius of the string constant, the the w(angular frequency) must change which is related to the period by the below equation which will also change,

$w = 2\pi /T$

to find out by how much the period will change we see that from the first equation that if we double the mass making it 2m then the w² will have to decrease by 2 that is it will become w²/2, at the same time keeping r constant since it says that in the question. We now absorb the 2 inside the square and we get,

$F_{c} = 2mr(w/\sqrt{2} )^{2} =2mr(2\pi /T\sqrt{2} )^{2}$

we can clearly see that the new period has become,

$T_{new} = \sqrt{2} T$

where T is the old period. So the new period is √2 times the old period given by the equation above.