The time period of an oscillating body is given by `T=2pisqrt((m)/(adg))`. What is the force equation for this body?

In physical pendulum, the time period for small oscillation is given by T=2pisqrt((I)/(Mgd)) where I is the moment of inertial of the body about an axis passing through a pivoted point O and perpendicular to the plane of oscillation and d is the separation point between centre of gravity and the pivoted point. In the physical pendulum a speacial point exists where if we concentrate the entire mass of body, the resulting simple pendulum (w.r.t. pivot point O) will have the same time period as that of physical pendulum This point is termed centre of oscillation. T=2pisqrt((I)/(Mgd))=2pisqrt((L)/(g)) Moreover, this point possesses two other important remarkable properties: Property I: Time period of physical pendulum about the centre of oscillation (if it would be pivoted) is same as in the original case. Property II: If an impulse is applied at the centre of oscillatioin in the plane of oscillation, the effect of this impulse at pivoted point is zero. Because of this property, this point is also known as the centre of percussion. From the given information answer the following question: Q. A uniform rod of mass M and length L is pivoted about point O as shown in Figgt It is slightly rotated from its mean position so that it performs angular simple harmonic motion. For this physical pendulum, determine the time period oscillation.

In physical pendulum, the time period for small oscillation is given by T=2pisqrt((I)/(Mgd)) where I is the moment of inertial of the body about an axis passing through a pivoted point O and perpendicular to the plane of oscillation and d is the separation point between centre of gravity and the pivoted point. In the physical pendulum a speacial point exists where if we concentrate the entire mass of body, the resulting simple pendulum (w.r.t. pivot point O) will have the same time period as that of physical pendulum This point is termed centre of oscillation. T=2pisqrt((I)/(Mgd))=2pisqrt((L)/(g)) Moreover, this point possesses two other important remarkable properties: Property I: Time period of physical pendulum about the centre of oscillation (if it would be pivoted) is same as in the original case. Property II: If an impulse is applied at the centre of oscillatioin in the plane of oscillation, the effect of this impulse at pivoted point is zero. Because of this property, this point is also known as the centre of percussion. From the given information answer the following question: Q. If an impulse J is applied at the centre of oscillation in the plane of oscillation, then angular velocity of the rod will be .

In physical pendulum, the time period for small oscillation is given by T=2pisqrt((I)/(Mgd)) where I is the moment of inertial of the body about an axis passing through a pivoted point O and perpendicular to the plane of oscillation and d is the separation point between centre of gravity and the pivoted point. In the physical pendulum a speacial point exists where if we concentrate the entire mass of body, the resulting simple pendulum (w.r.t. pivot point O) will have the same time period as that of physical pendulum This point is termed centre of oscillation. T=2pisqrt((I)/(Mgd))=2pisqrt((L)/(g)) Moreover, this point possesses two other important remarkable properties: Property I: Time period of physical pendulum about the centre of oscillation (if it would be pivoted) is same as in the original case. Property II: If an impulse is applied at the centre of oscillatioin in the plane of oscillation, the effect of this impulse at pivoted point is zero. Because of this property, this point is also known as the centre of percussion. From the given information answer the following question: Q. For the above question, locate the centre of oscillation.

The time period of oscillation of a body is given by T=2pisqrt((mgA)/(K)) K: Represents the kinetic energy, m mass, g acceleration due to gravity and A is unknown If [A]=M^(x)L^(y)T^(z) , then what is the value of x+y+z?

The time period of oscillation of simple pendulum is given by t = 2pisqrt(I//g) What is the accurancy in the determination of 'g' if 10cm length is knownj to 1mm accuracy and 0.5 s time period is measured form time of 100 oscillations with a wastch of 1 sec. resolution.

The periodic time (T) of a simple pendulum of length (L) is given by T=2pisqrt((L)/(g)) . What is the dimensional formula of Tsqrt((g)/(L)) ?

The period of oscillation of a simple pendulum is given by T=2pisqrt((l)/(g)) where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g is

The time period of oscillation of a simple pendulum is given by T=2pisqrt(l//g) The length of the pendulum is measured as 1=10+-0.1 cm and the time period as T=0.5+-0.02s . Determine percentage error in te value of g.

The time period 'T' of a simple pendulum of length 'l' is given by T = 2pisqrt(l/g) .Find the percentage error in the value of 'T' if the percentage error in the value of 'l' is 2%.