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The length of a simple pendulum is about 100 cm known to an accuracy of 1 mm. Its period of oscillation is 2s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution. What is the accuracy in the determined value of g ?
- a)0.2%
- b)0.5%
- c)0.1%
- d)2%
Correct answer is option 'A'. Can you explain this answer?
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Accuracy in Determining g using a Simple Pendulum
Given Data:
- Length of pendulum, L = 100 cm, with an accuracy of 1 mm
- Time period of oscillation, T = 2 s, with a clock resolution of 0.1 s
Formula:
The time period of a simple pendulum is given by:
T = 2π√(L/g)
where g is the acceleration due to gravity.
Calculations:
1. Let's first calculate the percentage uncertainty in the length of the pendulum:
Percentage uncertainty = (Accuracy/Measurement) x 100
= (1/1000) x 100 (since 1 mm = 0.1 cm)
= 0.1%
2. Now, let's calculate the percentage uncertainty in the time period:
Since we are measuring the time for 100 oscillations instead of one, the time period for one oscillation is:
t = T/100
= 0.02 s (since T = 2 s)
Percentage uncertainty = (Resolution/Measurement) x 100
= (0.1/0.02) x 100
= 0.5%
3. To calculate the uncertainty in g, we need to find the partial derivative of g with respect to L and T:
∂g/∂L = - (T^2/L^2) x (1/4π^2)
∂g/∂T = 2π^2L/T^3
4. Now, we can calculate the percentage uncertainty in g using the formula:
Percentage uncertainty in g = √((%uncertainty in L)^2 + (%uncertainty in T)^2 + 2(%uncertainty in L)(%uncertainty in T) x (∂g/∂L)(∂g/∂T)) x 100
Substituting the values:
Percentage uncertainty in g = √((0.1)^2 + (0.5)^2 + 2(0.1)(0.5) x (-0.015625) x 6.25) x 100
= 0.2%
Therefore, the accuracy in the determined value of g is 0.2%, which is option A.
Given Data:
- Length of pendulum, L = 100 cm, with an accuracy of 1 mm
- Time period of oscillation, T = 2 s, with a clock resolution of 0.1 s
Formula:
The time period of a simple pendulum is given by:
T = 2π√(L/g)
where g is the acceleration due to gravity.
Calculations:
1. Let's first calculate the percentage uncertainty in the length of the pendulum:
Percentage uncertainty = (Accuracy/Measurement) x 100
= (1/1000) x 100 (since 1 mm = 0.1 cm)
= 0.1%
2. Now, let's calculate the percentage uncertainty in the time period:
Since we are measuring the time for 100 oscillations instead of one, the time period for one oscillation is:
t = T/100
= 0.02 s (since T = 2 s)
Percentage uncertainty = (Resolution/Measurement) x 100
= (0.1/0.02) x 100
= 0.5%
3. To calculate the uncertainty in g, we need to find the partial derivative of g with respect to L and T:
∂g/∂L = - (T^2/L^2) x (1/4π^2)
∂g/∂T = 2π^2L/T^3
4. Now, we can calculate the percentage uncertainty in g using the formula:
Percentage uncertainty in g = √((%uncertainty in L)^2 + (%uncertainty in T)^2 + 2(%uncertainty in L)(%uncertainty in T) x (∂g/∂L)(∂g/∂T)) x 100
Substituting the values:
Percentage uncertainty in g = √((0.1)^2 + (0.5)^2 + 2(0.1)(0.5) x (-0.015625) x 6.25) x 100
= 0.2%
Therefore, the accuracy in the determined value of g is 0.2%, which is option A.
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The length of a simple pendulum is about 100 cm known to an accuracy of 1 mm. Its period of oscillation is 2s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution. What is the accuracy in the determined value of g ?a)0.2%b)0.5%c)0.1%d)2%Correct answer is option 'A'. Can you explain this answer?
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