The period of a simple pendulum is given by the formulaT = 2*pi*sqrt(L/g)
where T = period
L = length
and g = local acceleration due to gravity.
Note that this formula is applicable only when the angular displacement of the pendulum is small. For a displacement of 22.5 degrees (a quarter of a right angle), the true period is approx 1% longer : a clock will lose more than 1/2 a minute every hour!
the period T of a rigid-body compound pendulum for small angles is given byT=2π√I/mgRwhere I is the moment of inertia of the pendulum about the pivot point, m is the mass of the pendulum, and R is the distance between the pivot point and the center of mass of the pendulum.For example, for a pendulum made of a rigid uniform rod of length L pivoted at its end, I = (1/3)mL2. The center of mass is located in the center of the rod, so R = L/2. Substituting these values into the above equation gives T = 2π√2L/3g. This shows that a rigid rod pendulum has the same period as a simple pendulum of 2/3 its length.
The longer the length of the pendulum, the longer the time taken for the pendulum to complete 1 oscillation.
The most accurate way to model a pendulum (without air resistance) is as a differential equation in terms of the angle it makes with the vertical, θ, the length of the pendulum, l, and the acceleration due to gravity, g. d²θ/dt² = -g*sin(θ)/l There is no easy way to integrate this to get θ as a function of time, but if you assume θ is small, you can use the small angle approximation sin(θ)~θ which makes the equation d²θ/dt² = -g*θ/l Which can then be integrated to get the solution θ(t)=θmax*sin(t*√(g/l)) Using this equation, you can easily derive that the period of the pendulum (time required to go through one full cycle) would be T=2π*√(l/g) If air resistance is also accounted for in the original differential equation, the exact equation will be much harder to derive, but in general will involve an exponential decay of a sin function.
A longer pendulum will have a smaller frequency than a shorter pendulum.
To predict the period of a pendulum, we can use the equation T = 2Ļā(L/g), where T is the period, L is the length of the string, and g is the acceleration due to gravity. Plugging in L = 24cm (or 0.24m) and g = 9.8 m/sĀ², we can calculate the period using this equation.
The period of a pendulum is approximated by the equation T = 2 pi square-root (L / g). Note: This is only an approximation, applicable only for very small angles of swing. At larger angles, a circular error is introduced, but the basic equation still holds true.Looking at that equation, you see that time is proportional to the square root of the length of the pendulum, so to double the period of a pendulum you need to increase its length by a factor of four.
The period of oscillation of a simple pendulum displaced by a small angle is: T = (2*PI) * SquareRoot(L/g) where T is the period in seconds, L is the length of the string, and g is the gravitional field strength = 9.81 N/Kg. This equation is for a simple pendulum only. A simple pendulum is an idealised pendulum consisting of a point mass at the end of an inextensible, massless, frictionless string. You can use the simple pendulum model for any pendulum whose bob mass is much geater than the length of the string. For a physical (or real) pendulum: T = (2*PI) * SquareRoot( I/(mgr) ) where I is the moment of inertia, m is the mass of the centre of mass, g is the gravitational field strength and r is distance to the pivot from the centre of mass. This equation is for a pendulum whose mass is distributed not just at the bob, but throughout the pendulum. For example, a swinging plank of wood. If the pendulum resembles a point mass on the end of a string, then use the first equation.
The equation for the length, L, of a pendulum of time period, T, is gievn byL = g(T2/4?2),where g is the acceleration due to gravity. So, for a pendulum of time period 4.48 sec, the length of the pendulum is 4.99 metres (3 s.f).
the period T of a rigid-body compound pendulum for small angles is given byT=2π√I/mgRwhere I is the moment of inertia of the pendulum about the pivot point, m is the mass of the pendulum, and R is the distance between the pivot point and the center of mass of the pendulum.For example, for a pendulum made of a rigid uniform rod of length L pivoted at its end, I = (1/3)mL2. The center of mass is located in the center of the rod, so R = L/2. Substituting these values into the above equation gives T = 2π√2L/3g. This shows that a rigid rod pendulum has the same period as a simple pendulum of 2/3 its length.
Compound pendulum is a physical pendulum whereas a simple pendulum is ideal pendulum. The difference is that in simple pendulum centre of mass and centre of oscillation are at the same distance.
Without going through all the derivations, unless some one wants me to (I could show you my physics notes), the equation for a period of a pendulum with small amplitude (meaning reasonable amplitudes, i.e. less than 45O from the normal) is : T = 2 * Pi * sqrt(L / g) where L is the length of the pendulum g is the acceleration due to gravity where ever the pendulum is (9.8 m/s2 on earth)
A lift in free fall is the same as a lift with no gravity (e.g. in space), i.e. accelleration due to gravity, g = 0 ms^-2. Now your intuition should tell you what's going to happen but even if it doesn't you can plug this value into your equation for the pendulum's period to find out what happens.
Some of the classical mechanics for a slinky include The Klein Gordon Equation, Phase Velocity, Group Velocity, and The Sine-Gordon or Pendulum Equation. There is also Electrostatics, and The Discrete Fourier Transform.
A shorter pendulum has a shorter period. A longer pendulum has a longer period.
The longer the length of the pendulum, the longer the time taken for the pendulum to complete 1 oscillation.
maybe it is a pendulum
bifilar pendulum