rt is tied to the end of an unsireichedA toy cart isstring oF leng Lmo

1.	rt is tied to the end of an unsireichedA toy cart isstring oF leng Lmoves in horizontal circle co ith Tadidsr2」,andime period T. T it is stretehed Speeded untilit moves in horizon tel circle 아 rd dius '84,aimperiod Tn relation beth Tand-互is C Hookslaud obeyedcohen revolved the toy carJE IT2
Answer» The new time period is given by T₁ = √3T/2 Explanation: Case I string length = l string is stretched to 2l, so net increase in length = 2l - l = l => spring force, F = kl where k = spring constant when the string is rotated with 2l radius, we know that centripetal force is given by, F = mrω² = m x 2l x (2π/T)² Since the centripetal force and spring force should be equal, hence, kl = m x 2l x (2π/T)² => k = 2m x (2π/T)².............eq1 Case II string length = l string is stretched to 3l, so net increase in length = 3l - l = 2l => spring force, F = k x 2l = 2kl when the string is rotated with 3l radius, we know that centripetal force is given by, F = mrω₁² = m x 3l x (2π/T₁)² Since the centripetal force and spring force should be equal, hence, 2kl = m x 3l x (2π/T₁)² => k = 3m/2 x (2π/T₁)².............eq2 equating eq1 and eq2 we get: 2m x (2π/T)² = 3m/2 x (2π/T₁)² => T₁²/T² = 3/4 => T₁ = √3T/2 Hence the new time period is given by T₁ = √3T/2 hit like if you find it useful

rt is tied to the end of an unsireichedA toy cart isstring oF leng Lmoves in horizontal circle co ith Tadidsr2」,andime period T. T it is stretehed Speeded untilit moves in horizon tel circle 아 rd dius '84,aimperiod Tn relation beth Tand-互is C Hookslaud obeyedcohen revolved the toy carJE IT2

Answer»

The new time period is given by T₁ = √3T/2

Explanation:

Case I

string length = l

string is stretched to 2l, so net increase in length = 2l - l = l

=> spring force, F = kl

where k = spring constant

when the string is rotated with 2l radius,

we know that centripetal force is given by,

F = mrω² = m x 2l x (2π/T)²

Since the centripetal force and spring force should be equal, hence,

kl = m x 2l x (2π/T)²

=> k = 2m x (2π/T)².............eq1

Case II

string length = l

string is stretched to 3l, so net increase in length = 3l - l = 2l

=> spring force, F = k x 2l = 2kl

when the string is rotated with 3l radius,

we know that centripetal force is given by,

F = mrω₁² = m x 3l x (2π/T₁)²

Since the centripetal force and spring force should be equal, hence,

2kl = m x 3l x (2π/T₁)²

=> k = 3m/2 x (2π/T₁)².............eq2

equating eq1 and eq2 we get:

2m x (2π/T)² = 3m/2 x (2π/T₁)²

=> T₁²/T² = 3/4

=> T₁ = √3T/2

Hence the new time period is given by T₁ = √3T/2

hit like if you find it useful