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# two particles of mass m each are moving in a horizontal circle with same angular speed .if both string are of same length then the ratio of tension in the string T1/T2=?

Girija Sajjanar
22 Points
3 years ago
That answer is 1 since the mass, angular velocity and length of the string are same. Also the time period only depends on angular velocity as the formula for time period is 2pi/angular velocity(independent of mass and length of the string.

15 Points
2 years ago
$\dpi{200} \bg_white T =( Lcostheta/g)$1/2                                    here time period of conical pendulum is constant therefore                    T= T2       L1/L2 = cos theta2/cos theta                                                   nn                           = cos 45 /cos 30
=  (2/3) to the power 1/2 means root 2/3
Rajdeep
231 Points
2 years ago
HELLO THERE!

Things to consider here are that, mass of two bodies are same (m).
Length of both strings is same (l).
Angular velocity of both bodies is same (w).

Consider body first:
Since body 1 is in a horizontal circle with string of length l,
when you draw Free Body diagram of the body in motion,

we get the following two equations:

$Tcos\theta = mg$
$Tsin\theta = m\omega^{2}lsin\theta$

So, we get Tension:
$T_{1} = m\omega^{2}l$

Now, it is mentioned that mass of both bodies is same, angular velocity is same, and length of string is same: and Tension is calculated from these three parameters.

So, we conclude that Tensions in both strings are same:

$\frac{T_{1}}{T_{2}} = \frac{1}{1}$