Calculate the frequency of a simple pendulum of length 950 mm. Give answer to an appropriate number of significant figures.
Firstly, we remember that frequency is the reciprocal of the time period. The formula for the time period of a simple pendulum is T=2π*sqrt(l/g)
Secondly, we identify each part of the equation. We recall that 'l' is the length of the pendulum in metres (which in this case is 0.95 m) and 'g' is acceleration due to gravity (which has a magnitude of 9.81 m s-2).
Thirdly, we insert these values into their respected positions in the equation to give us T=2π*sqrt(0.95/9.81)=1.955 to four significant figures.
Fourthly, we take the reciprocal of this value to find frequency, as f=1/T. This gives us 0.5115 Hz to four significant figures.
Lastly, we put our answer in the "appropriate number of significant figures" asked for in the question. In this case, this is 3SF (because the lowest number of significant figures observed in the question is 3) and as a result, our answer is 0.512 Hz
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