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Question

Physics Question on simple harmonic motion

The period of oscillation of a simple pendulum is T=2πLgT=2 \pi \sqrt{\frac{L}{g}}. Measured value of LL is 20.0cm20.0\, cm known to 1mm1\, mm accuracy and time for 100100 oscillations of the pendulum is found to be 90s90\, s using a wrist watch of 1s1\, s resolution. The accuracy in the determination of gg is

A

3%

B

2%

C

1%

D

5%

Answer

3%

Explanation

Solution

dTT=12dLL12dgg\frac{d T}{T}=\frac{1}{2} \frac{d L}{L}-\frac{1}{2} \frac{d g}{g} 90100+1100\frac{90}{100}+\frac{1}{100} 12dgg=12dLLdTT\frac{1}{2} \frac{d g}{g}=\frac{1}{2} \frac{d L}{L}-\frac{d T}{T} 12×0.120=1/10090/100=1400+190\frac{1}{2} \times \frac{0.1}{20}=\frac{1 / 100}{90 / 100}=\frac{1}{400}+\frac{1}{90} 12dgg=1400+190\frac{1}{2} \frac{d g}{g}=\frac{1}{400}+\frac{1}{90} dgg=(490400×90)×2\frac{d g}{g}=\left(\frac{490}{400 \times 90}\right) \times 2 =(490200×90)=0.20272=\left(\frac{490}{200 \times 90}\right)=0.20272 =dg/g×1002.72%3%= dg / g \times 100 \approx 2.72 \% \approx 3 \%