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A student measures the time period …

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A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s,91s,92s & 95s. If the mimimum division in the measuring clock is 1s, then find the reported mean time ?

  • 1 answers

Payal Singh 6 years, 9 months ago

Let’s  first  find  the mean  value  of time:

{tex}t_{mean}={t_1+t_2+t_3+t_4\over 4}{/tex}

{tex}{90+91+95+92\over 4}= 92 {/tex}

Then,  we can  find  the  absolute error for each  measurement:

{tex}\Delta t_1=|t_{mean}-t_1|= |92-90|=2{/tex}

{tex}\Delta t_2=|t_{mean}-t_2|= |92-91|=1{/tex}

{tex}\Delta t_3=|t_{mean}-t_3|= |92-92|=0{/tex}

{tex}\Delta t_4=|t_{mean}-t_4|= |92-95|=3{/tex}

Let’s  calculate the mean  absolute error: 

{tex}\Delta t_{mean}={ \Delta t_1+ \Delta t_2+ \Delta t_3+ \Delta t_4\over 4}{/tex}

{tex}{2+1+0+3\over 4}= 1.5{/tex}{tex}\approx 2{/tex}

Therefore,  the reported  mean  time should  be  92 ± 2s

 

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