Yes.
The ideal pendulum consists of a massive bob suspended from a frictionless pivot by a massless string. For small angles the period is given by the formula: t = 2*pi*sqrt(l/g)
However, the formula depends on the assumption that, for the angle of displacement x (measured in radians), sin(x) approximately equals x. For large x the approximation does not hold true and so the formula needs amending.
For x = 0.4 radian, the period is about 1% greater than that given by the unadjusted formula.
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Height does not affect the period of a pendulum.
Yes. The derivation of the simple formula for the period of the pendulum requires the angle, theta (in radians) to be small so that sin(theta) and theta are approximately equal. There are more exact formulae, though.
The period of a pendulum is affected by the angle created by the swing of the pendulum, the length of the attachment to the mass, and the weight of the mass on the end of the pendulum.
The longer the pendulum is, the greater the period of each swing. If you increase the length four times, you will double the period. It is hard to notice, but the period of a pendulum does depend on the angle of oscillation. For small angles, the period is constant and depends only on the length of the pendulum. As the angle of oscillation (amplitude) is increased, additional factors of a Taylor approximation become important. (T=2*pi*sqrt(L/g)[1+theta^2/16+...] and the period increases. (see hyper physics: http://hyperphysics.phy-astr.gsu.edu/hbase/pendl.html) Interestingly, if the pendulum is supported by a very light wire then the mass of the object at the end of the pendulum does not affect the period. Obviously, the greater the mass, the less any air friction or friction at the pivot will slow the pendulum. Also interestingly, the pendulum period is dependant on the force of gravity on the object (g). One must not assume that g is constant for all places on Earth.
The period of the pendulum is unchanged by the angle of swing. See link.