A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/2. The time period of the second pendulum would be:
√2 T
Understanding Pendulum Time Period and Gravity
This question asks us to compare the time periods of two pendulum clocks under different gravitational conditions. The time period of a simple pendulum depends on its length and the acceleration due to gravity, but not on the mass of the bob.
Key Formula for Simple Pendulum
The time period (T) of a simple pendulum is given by the formula:
$$T = 2\pi \sqrt{\frac{L}{g}}$$
Where:
- \(L\) is the length of the pendulum.
- \(g\) is the acceleration due to gravity at the location of the pendulum.
Notice that the mass of the bob is not present in this formula, indicating that the time period is independent of the bob's mass.
Analyzing the First Pendulum Clock
For the first pendulum clock, let its length be \(L_1\), the mass of its bob be \(m_1\), and the gravitational acceleration be \(g_1\). Its time period is given as \(T\).
- Length, \(L_1 = L\)
- Mass, \(m_1 = m\)
- Gravitational acceleration, \(g_1 = g\)
- Time period, \(T_1 = T\)
Using the formula, we have:
$$T_1 = T = 2\pi \sqrt{\frac{L_1}{g_1}} = 2\pi \sqrt{\frac{L}{g}}$$
Analyzing the Second Pendulum Clock
For the second pendulum clock, its length is the same as the first, the mass of its bob is double, and the gravitational acceleration is half.
- Length, \(L_2 = L\) (same length as the first)
- Mass, \(m_2 = 2m\) (double the mass of the first)
- Gravitational acceleration, \(g_2 = g/2\) (half the value)
- Let its time period be \(T_2\)
Using the formula, the time period \(T_2\) is:
$$T_2 = 2\pi \sqrt{\frac{L_2}{g_2}}$$
Substitute the values \(L_2 = L\) and \(g_2 = g/2\):
$$T_2 = 2\pi \sqrt{\frac{L}{g/2}}$$
Calculating the Time Period of the Second Pendulum
Simplify the expression for \(T_2\):
$$T_2 = 2\pi \sqrt{\frac{L}{g/2}} = 2\pi \sqrt{\frac{L \times 2}{g}} = 2\pi \sqrt{\frac{2L}{g}}$$
We can separate the \(\sqrt{2}\) term:
$$T_2 = 2\pi \sqrt{2} \sqrt{\frac{L}{g}}$$
From the first pendulum, we know that \(T = 2\pi \sqrt{\frac{L}{g}}\). Substitute this into the equation for \(T_2\):
$$T_2 = \sqrt{2} \times \left(2\pi \sqrt{\frac{L}{g}}\right)$$
$$T_2 = \sqrt{2} T$$
Effect of Mass on Pendulum Time Period
It's important to note that while the mass of the bob for the second pendulum is double that of the first, this information is irrelevant to the time period calculation of an ideal simple pendulum. The formula \(T = 2\pi \sqrt{L/g}\) does not include mass.
Summary of Results
| Pendulum | Length (\(L\)) | Mass (\(m\)) | Gravity (\(g\)) | Time Period (\(T\)) |
|---|---|---|---|---|
| First | \(L\) | \(m\) | \(g\) | \(T\) |
| Second | \(L\) | \(2m\) | \(g/2\) | \(T_2\) |
Comparing the time periods:
$$T = 2\pi \sqrt{\frac{L}{g}}$$
$$T_2 = 2\pi \sqrt{\frac{L}{g/2}} = 2\pi \sqrt{\frac{2L}{g}} = \sqrt{2} \times 2\pi \sqrt{\frac{L}{g}} = \sqrt{2} T$$
Thus, the time period of the second pendulum is \(\sqrt{2}\) times the time period of the first pendulum.
Revision Table: Pendulum Clock Concepts
| Concept | Description | Formula | Factors Affecting Time Period |
|---|---|---|---|
| Simple Pendulum | An idealized point mass (bob) suspended by a massless, inextensible string from a fixed support. | N/A | Length, Acceleration due to Gravity |
| Time Period (T) | The time taken for one complete oscillation (swing back and forth). | \(T = 2\pi \sqrt{\frac{L}{g}}\) | Length (L), Gravity (g) |
| Frequency (f) | The number of oscillations per unit time. | \(f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{L}}\) | Length (L), Gravity (g) |
| Effect of Mass | For a simple pendulum, the mass of the bob does not affect the time period. | Mass (m) is not in formula | Independent of mass (m) |
| Effect of Amplitude | For small angles of oscillation (< about 10-15 degrees), the time period is nearly independent of amplitude. For large amplitudes, the period increases slightly. | Formula valid for small angles | Nearly independent for small angles |
Additional Information: Pendulum Clocks and Gravity
Pendulum clocks rely on the constant time period of a pendulum for accurate timekeeping. Because the time period depends on the local gravitational acceleration (\(g\)), a pendulum clock will run faster or slower if moved to a location with a different gravitational pull. For example:
- If \(g\) increases (e.g., moving from the equator towards the poles), the time period \(T\) decreases, and the clock runs faster.
- If \(g\) decreases (e.g., moving from the poles towards the equator, or going up a mountain), the time period \(T\) increases, and the clock runs slower.
The question describes lifting the clocks to different heights, which implies changes in the gravitational acceleration \(g\). Gravity decreases with altitude. The time period of a pendulum clock is directly proportional to the square root of its length and inversely proportional to the square root of the gravitational acceleration.
$$T \propto \sqrt{L}$$
$$T \propto \frac{1}{\sqrt{g}}$$
Understanding these relationships helps in predicting how a pendulum clock's timekeeping will change under varying conditions of length and gravity.
Important Questions from Simple Harmonic Motion
- The displacement of a particle is given by $y(t) = K + P \sin^2(\omega t) + Q \sin(\omega t) \cos(\omega t)$. If this represents a simple harmonic motion, the amplitude of its oscillation is:
- Which one of the following equations of motion represents simple harmonic motion?
Assume $A$, $B$, $C$, $D$, $m$, $k$, and $\omega$ are all positive constants. In simple harmonic motion, the particle velocity lags behind the displacement by a phase angle of __________.
- A particle undergoes simple harmonic motion. Determine the phase difference between its instantaneous velocity and instantaneous acceleration.
- A particle executes simple harmonic motion along a straight line. When its displacement from the mean position is $x$, its speed is $v$. If the displacement becomes $2x$, its speed reduces to $v/2$. What is the amplitude of the oscillation?
