The time period of a simple harmonic oscillator is 
. Measured value of mass 
of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant 
is 
.
Text Solution
Verified by ExpertsD
The given formula for time period is
.
Squaring both sides, we get 
, which gives 
.
The relative error in 
is given by 
.
Given: 
.
Total time for 
oscillations is 
with resolution 
.
Since 
, the relative error in 
is the same as in 
.
Substituting the values: 
.
Percentage error 
.
Rounding to two decimal places, we get 
.
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