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Simple Harmonic Motion question

2021 · 18 Mar · Shift 1 · Q71
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  5. /2021 · 18 Mar · Shift 1 · Q71

Simple Harmonic Motion question

2021 · 18 Mar · Shift 1 · Q71

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
The time period of a simple pendulum is given by T=2πlgT = 2\pi \sqrt {{l \over g}}T=2πgl​​. The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-
  1. A
    2%
  2. B
    3%
  3. C
    5%
  4. D
    4%
View written solutionFree

Correct answer: B

  1. Formula for ggg in terms of measured quantities

Given T=2πlgT = 2\pi\sqrt{\frac{l}{g}}T=2πgl​​ So, g=4π2lT2g = \frac{4\pi^2 l}{T^2}g=T24π2l​

Hence, the fractional error in ggg is Δgg=Δll+2ΔTT\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}gΔg​=lΔl​+2TΔT​

  1. Error in length measurement

Length of pendulum: l=10 cm=100 mml = 10\text{ cm} = 100\text{ mm}l=10 cm=100 mm Accuracy in length measurement = 1 mm1\text{ mm}1 mm

Therefore, Δll=1100=0.01=1%\frac{\Delta l}{l} = \frac{1}{100} = 0.01 = 1\%lΔl​=1001​=0.01=1%

  1. Error in time period measurement

Time for 200 oscillations: t=100 st = 100\text{ s}t=100 s Clock resolution = 1 s1\text{ s}1 s So error in measured total time is Δt=1 s\Delta t = 1\text{ s}Δt=1 s

Since T=t200T = \frac{t}{200}T=200t​ we have ΔTT=Δtt=1100=0.01=1%\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{100} = 0.01 = 1\%TΔT​=tΔt​=1001​=0.01=1%

  1. Calculate percentage error in ggg

Using Δgg=Δll+2ΔTT\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}gΔg​=lΔl​+2TΔT​

Δgg=1%+2(1%)=3%\frac{\Delta g}{g} = 1\% + 2(1\%) = 3\%gΔg​=1%+2(1%)=3%

So, x=3%x = 3\%x=3%

Nearest integer = 3.

  1. Option check
  • A: 2%2\%2% ❌
  • B: 3%3\%3% ✅
  • C: 5%5\%5% ❌
  • D: 4%4\%4% ❌

Therefore, the correct option is B.

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