General Principles in the Working of Simple Pendulum, Simple Pulleys, Siphon
General Principles in the Working of Simple Pendulum, Simple Pulleys, Siphon MCQ Questions
13.
The restoring force in a simple pendulum is directly proportional to which quantity?
D.
Amplitude of oscillation
ANSWER :
B. Displacement angle
Approach
To determine the quantity to which the restoring force in a simple pendulum is directly proportional, we need to analyze the forces acting on the pendulum bob and derive the expression for the restoring force. The restoring force is the component of gravity that acts to bring the pendulum back to its equilibrium position.
Step-by-step
- Identify forces: When a simple pendulum of mass \(m\) is displaced by an angle \(\theta\) from its vertical equilibrium position, two main forces act on the bob: the tension \(T\) in the string and the gravitational force \(mg\) acting vertically downwards.
- Resolve gravitational force: We resolve the gravitational force \(mg\) into two components: one along the string and one perpendicular to the string.
- The component along the string is \(mg \cos\theta\), which is balanced by the tension \(T\) (approximately, for small angles).
- The component perpendicular to the string is \(mg \sin\theta\). This component always acts towards the equilibrium position and is responsible for bringing the pendulum back. This is the restoring force.
- Define restoring force: The restoring force \(F_r\) is given by:\[ F_r = -mg \sin\theta \]The negative sign indicates that the restoring force acts in the opposite direction to the displacement.
- Small angle approximation: For a simple pendulum, oscillations are considered simple harmonic motion (SHM) only for small angles of displacement. For small angles (typically \(\theta < 10^\circ\)), \(\sin\theta \approx \theta\) (where \(\theta\) is in radians).
- Substitute approximation: Substituting this approximation into the restoring force equation:\[ F_r \approx -mg \theta \]
- Conclusion: From this equation, it is clear that for small angles, the restoring force \(F_r\) is directly proportional to the displacement angle \(\theta\).
Correct Option
B) Displacement angle — As derived, for small angles of oscillation, the restoring force \(F_r = -mg \sin\theta \approx -mg \theta\). This shows that the restoring force is directly proportional to the displacement angle \(\theta\).
Incorrect Options
- A) Mass of the bob — While the restoring force does depend on the mass of the bob (\(F_r = -mg \sin\theta\)), it is not the quantity to which the restoring force is directly proportional in the context of the pendulum's motion being restored to equilibrium based on its displacement. The question asks what it is proportional to, and the displacement angle is the primary variable that determines the magnitude of the restoring force for a given pendulum.
- C) Length of the string — The length of the string (\(L\)) affects the period of oscillation (\(T = 2\pi \sqrt{L/g}\)) and the arc length of displacement (\(x = L\theta\)), but the restoring force itself is fundamentally proportional to \(\sin\theta\) (or \(\theta\) for small angles), not directly to \(L\).
- D) Amplitude of oscillation — The amplitude is the maximum displacement angle. While the maximum restoring force occurs at the amplitude, the restoring force at any given instant is proportional to the instantaneous displacement angle, not the amplitude itself. The amplitude is a constant for a particular oscillation, whereas the restoring force varies with the instantaneous displacement.
14.
In a simple pendulum, what is the relationship between time period and acceleration due to gravity?
A.
Time period is directly proportional to gravity
B.
Time period is inversely proportional to gravity
C.
Time period is independent of gravity
D.
Time period is equal to gravity
ANSWER :
B. Time period is inversely proportional to gravity
Approach
The question asks about the relationship between the time period of a simple pendulum and the acceleration due to gravity. To answer this, we need to recall the formula for the time period of a simple pendulum.
Step-by-step
- Recall the formula for the time period of a 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 and \(g\) is the acceleration due to gravity.
- Analyze the relationship from the formula: From the formula, we can observe how \(T\) depends on \(g\). The term \(g\) is in the denominator under the square root sign.
- Determine the proportionality: This means that \(T\) is inversely proportional to the square root of \(g\). If \(g\) increases, \(T\) decreases, and if \(g\) decreases, \(T\) increases. Therefore, the time period is inversely proportional to the acceleration due to gravity.
Correct Option:
B) Time period is inversely proportional to gravity — As derived from the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), the time period \(T\) is inversely proportional to the square root of the acceleration due to gravity \(g\). This means that as \(g\) increases, \(T\) decreases, and vice versa.
Incorrect Options
- A) Time period is directly proportional to gravity — This is incorrect. If it were directly proportional, \(g\) would be in the numerator, which is not the case.
- C) Time period is independent of gravity — This is incorrect. The formula clearly shows \(g\) as a variable influencing \(T\). If \(g\) changes (e.g., on the moon vs. Earth), the time period will change.
- D) Time period is equal to gravity — This is incorrect. Time period is measured in seconds, and gravity (acceleration due to gravity) is measured in meters per second squared. They are different physical quantities and cannot be equal.
15.
What is the angle of displacement in a simple pendulum for small amplitude oscillations?
C.
Greater than 45 degrees
ANSWER :
A. Less than 5 degrees
Approach
To understand the angle of displacement for small amplitude oscillations in a simple pendulum, we need to recall the conditions under which the simple pendulum approximates Simple Harmonic Motion (SHM). This approximation is crucial for the standard formula for the time period of a simple pendulum.
Step-by-step
-
The equation of motion for a simple pendulum is given by:
\[ \frac{d^2\theta}{dt^2} + \frac{g}{L} \sin\theta = 0 \]
where \(\theta\) is the angular displacement, \(g\) is the acceleration due to gravity, and \(L\) is the length of the pendulum.
-
For the pendulum to execute Simple Harmonic Motion (SHM), the restoring force (or torque) must be directly proportional to the displacement and directed towards the equilibrium position. In the case of a simple pendulum, this means the term \(\sin\theta\) must be approximately equal to \(\theta\) (in radians).
-
The approximation \(\sin\theta \approx \theta\) is valid only for very small angles. When \(\theta\) is small, the Taylor series expansion of \(\sin\theta\) is:
\[ \sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots \]
For small angles, we can neglect higher-order terms, leading to \(\sin\theta \approx \theta\).
-
If this approximation is made, the equation of motion becomes:
\[ \frac{d^2\theta}{dt^2} + \frac{g}{L} \theta = 0 \]
This is the standard differential equation for SHM, from which the time period \(T = 2\pi \sqrt{\frac{L}{g}}\) is derived.
-
Practically, the approximation \(\sin\theta \approx \theta\) holds well for angles less than about 10-15 degrees. However, for a more accurate and ideal SHM, the angle is typically considered to be much smaller, usually less than 5 degrees (or even 2-3 degrees in some contexts) to minimize the error introduced by the approximation.
-
For example, at \(\theta = 5^\circ\) (which is approximately \(0.08727\) radians), \(\sin(5^\circ) \approx 0.08716\). The percentage difference is very small:
\[ \frac{|\theta - \sin\theta|}{\theta} \times 100\% = \frac{|0.08727 - 0.08716|}{0.08727} \times 100\% \approx 0.12\% \]
This small error ensures that the motion closely approximates SHM.
Correct Option: A) Less than 5 degrees
For a simple pendulum to exhibit small amplitude oscillations, which approximate Simple Harmonic Motion (SHM
16.
How does damping affect the motion of a simple pendulum?
D.
Removes oscillations completely
ANSWER :
B. Decreases amplitude
Approach
The question asks about the effect of damping on the motion of a simple pendulum. Damping refers to any effect that dissipates energy from an oscillating system. We need to understand how this energy loss manifests in the pendulum's motion.
Step-by-step
- Understanding Simple Harmonic Motion (SHM): A simple pendulum, in an ideal scenario (no air resistance, no friction at the pivot), undergoes Simple Harmonic Motion (SHM). In SHM, the total mechanical energy (potential + kinetic) remains constant, and the oscillation continues indefinitely with a constant amplitude.
- Introducing Damping: Damping forces are resistive forces that oppose motion and convert mechanical energy into other forms, typically heat. Examples include air resistance, friction at the pivot, and internal friction within the string.
- Effect of Damping on Energy: When damping is present, the pendulum continuously loses mechanical energy to its surroundings. This energy loss means that the pendulum cannot reach the same maximum height (potential energy) or maximum speed (kinetic energy) in successive oscillations.
- Impact on Amplitude: The amplitude of an oscillation is the maximum displacement from the equilibrium position. Since energy is being lost, the pendulum cannot swing as far from its equilibrium position in each subsequent swing. Therefore, the maximum displacement, which is the amplitude, progressively decreases over time.
- Impact on Period/Frequency: While damping primarily affects amplitude, it can also slightly increase the period (and thus decrease the frequency) of oscillation, especially for heavy damping. However, the most prominent and direct effect is on the amplitude.
Correct Option:
B) Decreases amplitude
Damping forces dissipate energy from the oscillating system. As the total mechanical energy of the pendulum decreases, its ability to swing to the same maximum displacement (amplitude) is reduced. Consequently, the amplitude of oscillation progressively decreases over time.
Incorrect Options
- A) Increases amplitude — This is incorrect. Increasing amplitude would imply an increase in the pendulum's energy, which is contrary to the definition of damping.
- C) Makes motion faster — This is generally incorrect. Damping forces oppose motion, which tends to slow down the pendulum's instantaneous speed, especially as it approaches its maximum displacement. While the period might slightly increase (making it slower in terms of completing an oscillation), the primary effect is not to make the motion faster.
- D) Removes oscillations completely — While damping eventually leads to the cessation of oscillations (the pendulum comes to rest at its equilibrium position), it doesn't remove them completely instantaneously. It causes the oscillations to gradually die out over time, meaning the amplitude decreases until it reaches zero. The statement implies an immediate or abrupt stop, which is not how damping typically works in a simple pendulum.
17.
Which factor remains constant in the oscillatory motion of a simple pendulum?
Approach
The question asks about a factor that remains constant in the oscillatory motion of a simple pendulum. This relates to the principle of conservation of mechanical energy, especially in an ideal scenario where there is no air resistance or friction.
Step-by-step
- Understanding Simple Pendulum Motion: A simple pendulum consists of a point mass (bob) suspended by a light inextensible string from a fixed support. When displaced from its equilibrium position and released, it oscillates due to gravity.
- Energy Transformation: During its oscillation, the pendulum continuously converts potential energy into kinetic energy and vice versa.
- At the extreme positions (maximum displacement), the pendulum momentarily stops, meaning its kinetic energy is zero, and its potential energy is maximum.
- At the mean position (equilibrium position), the pendulum has maximum speed, meaning its kinetic energy is maximum, and its potential energy is minimum (often considered zero at this point).
- Conservation of Mechanical Energy: In an ideal simple pendulum (i.e., neglecting air resistance and friction), the total mechanical energy, which is the sum of its kinetic energy and potential energy, remains constant throughout its motion. This is a fundamental principle of physics.
- Mathematical Representation:
Total Mechanical Energy \(E_{\text{total}} = \text{Kinetic Energy} (KE) + \text{Potential Energy} (PE)\)
Where:
\[ KE = \frac{1}{2}mv^2 \]\[ PE = mgh \] Here, \(m\) is the mass of the bob, \(v\) is its speed, \(g\) is the acceleration due to gravity, and \(h\) is its height above a reference point. As \(v\) and \(h\) change, \(KE\) and \(PE\) change, but their sum \(E_{\text{total}}\) remains constant.
Correct Option: B) Total energy
In the oscillatory motion of a simple pendulum, assuming no energy loss due to air resistance or friction, the total mechanical energy (sum of kinetic and potential energy) remains constant. Energy is continuously converted between kinetic and potential forms, but their sum stays the same.
Incorrect Options
- A) Speed — Speed is not constant. It is maximum at the mean position and zero at the extreme positions.
- C) Kinetic energy — Kinetic energy is not constant. It is maximum at the mean position and zero at the extreme positions.
- D) Displacement — Displacement is not constant. It varies from zero at the mean position to a maximum value at the extreme positions.
18.
What happens to the time period if the length of the pendulum is halved?
B.
Time period remains constant
ANSWER :
C. Time period decreases
Approach
The time period of a simple pendulum depends on its length and the acceleration due to gravity. We need to use the formula for the time period of a simple pendulum to determine how it changes when the length is halved.
Step-by-step
The formula for the time period (\(T\)) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \] where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
Let the initial time period be \(T_1\) and the initial length be \(L_1\). So,
\[ T_1 = 2\pi \sqrt{\frac{L_1}{g}} \]Now, the length of the pendulum is halved. So, the new length \(L_2\) will be \(L_2 = \frac{L_1}{2}\).
Let the new time period be \(T_2\). Substituting the new length into the formula:
\[ T_2 = 2\pi \sqrt{\frac{L_2}{g}} = 2\pi \sqrt{\frac{L_1/2}{g}} = 2\pi \sqrt{\frac{1}{2} \cdot \frac{L_1}{g}} \]We can separate the square root:
\[ T_2 = 2\pi \sqrt{\frac{1}{2}} \sqrt{\frac{L_1}{g}} \]Since \[\sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\], we get:
\[ T_2 = \frac{1}{\sqrt{2}} \left( 2\pi \sqrt{\frac{L_1}{g}} \right) \]We know that \(T_1 = 2\pi \sqrt{\frac{L_1}{g}}\). Therefore,
\[ T_2 = \frac{1}{\sqrt{2}} T_1 \]Since \(\sqrt{2} \approx 1.414\), \(T_2 \approx \frac{1}{1.414} T_1 \approx 0.707 T_1\). This means the new time period is approximately 70.7% of the original time period.
As \(0.707 < 1\), the time period decreases.
Correct Option:
C) Time period decreases. When the length of the pendulum is halved, the time period decreases by a factor of \(1/\sqrt{2}\).
Incorrect Options
- A) Time period doubles — This would happen if the length were quadrupled, as \(T \propto \sqrt{L}\).
- B) Time period remains constant — The time period is directly dependent on the length, so it cannot remain constant if the length changes.
- D) Time period increases — The time period increases if the length increases, not if it is halved.
