Question

The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum of 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 __________ %.

Hint:

In error analysis problems, the first step is always to find the correct physical formula relating the quantities. Once you have an expression like \( Z \propto X^a Y^b \), the percentage error in Z is simply \( |a|(% \text{error in } X) + |b|(% \text{error in } Y) \).

Answer 1

Correct Answer: 14

Step 1: Understanding the Question:
We are given the percentage errors in the measurement of acceleration due to gravity (g) and the time period (T) of a simple pendulum. We need to find the percentage error in the calculated energy (E) of the pendulum's oscillation.
Step 2: Key Formula or Approach:
1. We need a formula for the energy of a simple pendulum `E` in terms of `g` and `T`.
2. The total energy of a simple pendulum for small oscillations is \( E = \frac{1}{2} m \omega^2 A^2 \), where A is the amplitude. The amplitude can be expressed as \( A = L \theta_0 \) for a small angular amplitude \(\theta_0\).
3. The angular frequency is \( \omega = \sqrt{g/L} \). The time period is \( T = 2\pi \sqrt{L/g} \).
4. From the time period formula, we can express the length `L` as \( L = \frac{gT^2}{4\pi^2} \).
5. By substituting these relationships, we can find how `E` depends on `g` and `T`.
6. For a quantity \( Z = k \cdot g^a T^b \), the percentage error is given by \( \frac{\Delta Z}{Z} \times 100 = |a| \left(\frac{\Delta g}{g} \times 100\right) + |b| \left(\frac{\Delta T}{T} \times 100\right) \).
Step 3: Detailed Explanation:
Let's establish the relationship between E, g, and T. The energy of oscillation is proportional to the potential energy at maximum displacement, \( E \propto mg h \). For a pendulum of length `L` and angular amplitude \(\theta_0\), the maximum height `h` is \( h = L(1-\cos\theta_0) \approx L\frac{\theta_0^2}{2} \) for small angles. So, \( E \propto mgL \). (Assuming \(\theta_0\) is a fixed parameter).
Now we express `L` in terms of `g` and `T`.
From \( T = 2\pi \sqrt{\frac{L}{g}} \), we get \( T^2 = 4\pi^2 \frac{L}{g} \), which gives \( L = \frac{gT^2}{4\pi^2} \).
Substitute this expression for `L` into our energy relation:
\[ E \propto mgL \propto mg \left( \frac{gT^2}{4\pi^2} \right) \] \[ E \propto g^2 T^2 \] (Since `m` and \(4\pi^2\) are constants).
Now we can find the percentage error in E using the error propagation formula.
\[ \frac{\Delta E}{E} \times 100 = 2 \left(\frac{\Delta g}{g} \times 100\right) + 2 \left(\frac{\Delta T}{T} \times 100\right) \] We are given:
Percentage accuracy in g, \( \frac{\Delta g}{g} \times 100 = 4% \).
Percentage accuracy in T, \( \frac{\Delta T}{T} \times 100 = 3% \).
Substitute these values:
\[ \frac{\Delta E}{E} \times 100 = 2(4%) + 2(3%) = 8% + 6% = 14% \] Step 4: Final Answer:
The accuracy to which E is known is 14%.