Question

The time period of a spring block system is given as T=2π√m/k where m is mass and k is spring constant if T=(20.0±0.1) and m= (100±1)g Find the maximum percentage error in calculation of K?

Answer 1

Answer:

hy mate

Explanation:

The period of a mass m on a spring of spring constant k can be calculated as T=2π√mk T = 2 π m k ....refer the attachment

Answer 2

Answer:

The percentage error in calculation of K is 2%

Explanation:

Given that,

The time period for spring block mass system is

$T=2\pi \: \sqrt{ \frac{m}{k} }$

Squaring on both sides we get

$T {}^{2} =4π {}^{2} \frac{m}{k} \\ = > k = 4\pi {}^{2} \frac{m}{T {}^{2} }$

Applying the error derivative on both the sides we have

$\frac{∆k}{k} = \frac{∆m}{m} + 2( \frac{∆T}{T} )$

According to given data we have

Mass(m)=100g,∆m=1g

Time period(T)=20s,∆T=0.1s

Substituting these values in the above relation we get

$\frac{∆k}{k} = \frac{1}{100} + 2( \frac{0.1}{20} ) \\ = \frac{1}{100} + \frac{1}{100} = \frac{2}{100} = 0.02$

Hence percentage error in calculation of K is

$0.02 \times 100 = 2$

Therefore,The percentage error in calculation of K is 2%

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