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A mathematical pendulum 1 m long is deflected from the vertical by an angle of 40^(@)and let go. Find the period of oscillations using numerical methods. What will be the error, if in this case we use the formula for small oscillations? |
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Answer» `v-sqrt(2gh)=sqrt(2gl(cosalpha-cosalpha_(0)))=sqrt(2gl)` where `x=sqrt(cosalpha-cosalpha_(0))`. The body travels along the element of arc `Deltas=lDeltaalpha` in time `Deltal=(Deltas)/(v_(nv))=(lDeltaalpha)/(x_(AV)sqrt(2gl))=(Deltaalpha)/(x_(n)sqrt(2))sqrt((l)/(g))` PUT `Deltaalpha=3^(@)=pi//60` radian. Then `Deltat=2pisqrt((1)/(g)).(sqrt2)/(240x_(av))` Theperiod of oscillations is `T=4(Deltat_(1)+Deltat_(2)+......)=2pisqrt((l)/(g))(sqrt2)/(60)((1)/(x_(1av))+(1)/(x_(2av))+)`  Since the formula `T_(0)=2pisqrt(l//g)` is used to compute the period of small oscillations, it follows that `T=kT_(0)`, where `k=(sqrt2)/(60)((1)/(x_(1av))+(1)/(x_(2av))+.........)` is the correction factor. Let us compile a table from the COMPUTED data. We have `k=(sqrt2xx44.23)/(60)=1.042` Hence, the period `T=2piksqrt(l//g)=2.08` s, the VALUE of `T_(0)` being `T_(0)2pisqrt(l//g)=2.00` s. In this case the relative error DUE to the use of the formula for small oscillations will be 4%. |
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