What happens to the constants like $8\pi^2$ and $m$?

Constants have exact, known values, meaning their uncertainty or error is zero (e.g., $\Delta m = 0$, $\Delta \pi = 0$). Therefore, they do not contribute to the relative error of the final calculated quantity.

JEE Main 2021

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Animated Solution for Physics - Physics and Measurement: The acceleration due to gravity is found upto an accuracy of 4\% on a planet. The energy supplied to a simple pendulum to known mass $m$ to undertake oscillations of time period $T$ is being estimated. If time period is measured to an accuracy of 3\%, the accuracy to which $E$ is known as ..........\%.

Q: Why did we substitute $L$ instead of just finding the error in $L$ first?

You could absolutely find the error in $L$ first! Since $L \propto g T^2$, the error in $L$ is $\frac{\Delta L}{L} = \frac{\Delta g}{g} + 2\frac{\Delta T}{T} = 4\% + 2(3\%) = 10\%$. Then, since $E \propto g L$, the error in $E$ is $\frac{\Delta g}{g} + \frac{\Delta L}{L} = 4\% + 10\% = 14\%$. Substituting first just combines it into a single, elegant step.

Q: Why do the powers become multipliers in the error formula?

This comes from logarithmic differentiation. If $y = x^n$, taking the natural log gives $\ln y = n \ln x$. Differentiating both sides yields $\frac{dy}{y} = n \frac{dx}{x}$. For maximum possible error, we take the absolute values, so the power $n$ becomes a direct multiplier.

Enter Numerical Value:

Visualized Solution

$E = \frac{1}{2} m g L θ^{2}$

The energy of a simple pendulum is given by:
$E = \frac{1}{2} m g L θ^{2}$

$T = 2 π \frac{L}{g}$

The time period of a simple pendulum is:
$T = 2 π \frac{L}{g}$

$L = \frac{g T ^{2}}{4 π ^{2}}$

Expressing length $L$ in terms of $T$ and $g$ :
$L = \frac{g T ^{2}}{4 π ^{2}}$

$E = \frac{1}{2} m g (\frac{g T ^{2}}{4 π ^{2}}) θ^{2}$

Substituting $L$ back into the energy equation:
$E = \frac{1}{2} m g (\frac{g T ^{2}}{4 π ^{2}}) θ^{2}$

$E = \frac{m g ^{2} T ^{2} θ ^{2}}{8 π ^{2}}$

Simplifying the energy expression:
$E = \frac{m g ^{2} T ^{2} θ ^{2}}{8 π ^{2}}$

$\frac{Δ E}{E} = 2 \frac{Δ g}{g} + 2 \frac{Δ T}{T}$

Applying rules of relative error:
$\frac{Δ E}{E} = 2 \frac{Δ g}{g} + 2 \frac{Δ T}{T}$

$\frac{Δ E}{E} \times 100 = 2 (4%) + 2 (3%)$

Substituting the given percentage errors:
$\frac{Δ g}{g} \times 100 = 4%$
$\frac{Δ T}{T} \times 100 = 3%$
$\frac{Δ E}{E} \times 100 = 2 (4%) + 2 (3%)$

$\frac{Δ E}{E} \times 100 = 14%$

Final calculation:
$\frac{Δ E}{E} \times 100 = 8% + 6% = 14%$

$Accuracy = 14%$

The accuracy to which $E$ is known is $14%$ .

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The Sigma Insight: Errors in Measurement

Solution Diagram

The Energy of a Pendulum

Imagine you are setting up a simple pendulum in a laboratory. To keep it oscillating with a specific angular amplitude

θ

, you must supply it with a certain amount of energy. The expression for this total energy is given by:

E = \frac{1}{2} m g L θ^{2}

Here,

m

is the mass of the bob,

g

is the acceleration due to gravity,

L

is the length of the string, and

θ

is the angular amplitude.

Eliminating the Unknowns

Now, look closely at the problem statement. We are asked to find the percentage error in the energy

E

. We are given the percentage errors for gravity

g

(

4%

) and the time period

T

(

3%

). However, we have a slight issue: our energy equation contains the length

L

, but we have absolutely no information about the error in

L

To solve this, we must eliminate

L

by expressing it in terms of the variables we do know. We can use the standard formula for the time period of a simple pendulum:

T = 2 π \frac{L}{g}

By squaring both sides and rearranging the terms, we can isolate

L

L = \frac{g T ^{2}}{4 π ^{2}}

Next, we take this expression for

L

and substitute it back into our original energy equation. This is a crucial step because it transforms our energy equation into the variables we actually have data for:

E = \frac{1}{2} m g (\frac{g T ^{2}}{4 π ^{2}}) θ^{2}

Simplifying this, we get our master equation for error analysis:

E = \frac{m g ^{2} T ^{2} θ ^{2}}{8 π ^{2}}

The Power of Error Propagation

Now for the magic of error analysis! When calculating the maximum relative error, constants like mass

m

, amplitude

θ

, and

8 π^{2}

drop out because they have zero uncertainty (they are assumed to be exact or perfectly known in this context).

The rules of error propagation state that the exponents of our variables simply come down as multipliers. Therefore, the relative error in

E

is given by:

\frac{Δ E}{E} = 2 \frac{Δ g}{g} + 2 \frac{Δ T}{T}

Final Calculation

The problem states that the percentage accuracy in gravity is

4%

, and in the time period is

3%

. Let's carefully plug these given values into our relative error equation:

\frac{Δ E}{E} \times 100 = 2 (4%) + 2 (3%)

Let's do the final math. Two times four percent gives us

8%

, and two times three percent gives us

6%

. Adding them together, we get:

\frac{Δ E}{E} \times 100 = 8% + 6% = 14%

The accuracy to which the energy $E$ is known is $14%$ .

Common Traps & FAQs

Why do we use the formula

E = \frac{1}{2} m g L θ^{2}

The total energy of a simple pendulum is equal to its maximum potential energy, which is

m g h

. For small angles, the height

h

is approximately

L (1 - cos θ) \approx L (\frac{θ ^{2}}{2})

. Thus, substituting this gives

E = \frac{1}{2} m g L θ^{2}

Why did we substitute

L

instead of just finding the error in

L

first?

You could absolutely find the error in

L

first! Since

L \propto g T^{2}

, the error in

L

\frac{Δ L}{L} = \frac{Δ g}{g} + 2 \frac{Δ T}{T} = 4% + 2 (3%) = 10%

. Then, since

E \propto g L

, the error in

E

\frac{Δ g}{g} + \frac{Δ L}{L} = 4% + 10% = 14%

. Substituting first just combines it into a single, elegant step.

Why do the powers become multipliers in the error formula?

This comes from logarithmic differentiation. If

y = x^{n}

, taking the natural log gives

ln y = n ln x

. Differentiating both sides yields

\frac{d y}{y} = n \frac{d x}{x}

. For maximum possible error, we take the absolute values, so the power

n

becomes a direct multiplier.

What happens to the constants like

8 π^{2}

and

m

Constants have exact, known values, meaning their uncertainty or error is zero (e.g.,

Δ m = 0

Δ π = 0

). Therefore, they do not contribute to the relative error of the final calculated quantity.

Why do we add the errors instead of subtracting them?

When calculating the maximum possible percentage error, we always add the absolute values of the individual relative errors. This ensures we account for the worst-case scenario where errors compound rather than cancel each other out.

Enter Numerical Value:

Visualized Solution

$E = \frac{1}{2} m g L θ^{2}$

$T = 2 π \frac{L}{g}$

$L = \frac{g T ^{2}}{4 π ^{2}}$

$E = \frac{1}{2} m g (\frac{g T ^{2}}{4 π ^{2}}) θ^{2}$

$E = \frac{m g ^{2} T ^{2} θ ^{2}}{8 π ^{2}}$

$\frac{Δ E}{E} = 2 \frac{Δ g}{g} + 2 \frac{Δ T}{T}$

$\frac{Δ E}{E} \times 100 = 2 (4%) + 2 (3%)$

$\frac{Δ E}{E} \times 100 = 14%$

$Accuracy = 14%$

The Sigma Insight: Errors in Measurement

The Energy of a Pendulum

Eliminating the Unknowns

The Power of Error Propagation

Final Calculation

Similar Questions

A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0 cm$ and a stop watch with $1 s$ resolution measures the time taken for $40$ oscillations to be $50 s$ . The accuracy in $g$ is

A student uses a simple pendulum of exactly $1 m$ length to determine $g$ , the acceleration due to gravity. He uses a stop watch with the least count of $1 s$ for this and records $40 s$ for $20$ oscillations. For this observation, which of the following statement(s) is/are true?

The period of oscillation of a simple pendulum is $T = 2 π \frac{L}{g}$ . Measured value of $L$ is $20.0 cm$ known to $1 mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90 s$ using a wrist watch of $1 s$ resolution. The accuracy in the determination of $g$ is

The energy of a system as a function of time $t$ is given as $E (t) = A^{2} exp (- α t)$ , where $α = 0.2 s^{- 1}$ . The measurement of $A$ has an error of $1.25%$ . If the error in the measurement of time is $1.50%$ , the percentage error in the value of $E (t)$ at $t = 5 s$ is

If the length of the pendulum in pendulum clock increases by $0.1%$ , then the error in time per day is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=21​mgLθ2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=2πgL​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L=4π2gT2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=21​mg(4π2gT2​)θ2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=8π2mg2T2θ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EΔE​=2gΔg​+2TΔT​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EΔE​×100=2(4%)+2(3%)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EΔE​×100=14%

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Accuracy=14%

The Sigma Insight: Errors in Measurement

The Energy of a Pendulum

Eliminating the Unknowns

The Power of Error Propagation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is

$E = \frac{1}{2} m g L θ^{2}$

$T = 2 π \frac{L}{g}$

$L = \frac{g T ^{2}}{4 π ^{2}}$

$E = \frac{1}{2} m g (\frac{g T ^{2}}{4 π ^{2}}) θ^{2}$

$E = \frac{m g ^{2} T ^{2} θ ^{2}}{8 π ^{2}}$

$\frac{Δ E}{E} = 2 \frac{Δ g}{g} + 2 \frac{Δ T}{T}$

$\frac{Δ E}{E} \times 100 = 2 (4%) + 2 (3%)$

$\frac{Δ E}{E} \times 100 = 14%$

$Accuracy = 14%$

If the length of the pendulum in pendulum clock increases by $0.1%$ , then the error in time per day is