The problem asks about the change in the time period of a swing when a child stands up from a sitting position. This involves understanding the concept of a simple pendulum and how its time period is affected by the effective length and moment of inertia.
Understand the swing as a pendulum: A swing, with a person on it, can be approximated as a physical pendulum. The time period \(T\) of a physical pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgL}} \]where \(I\) is the moment of inertia about the pivot point, \(m\) is the total mass, \(g\) is the acceleration due to gravity, and \(L\) is the distance from the pivot point to the center of mass.
Analyze the change in position: When the child stands up from a sitting position on the swing, their center of mass shifts upwards. This means the distance \(L\) (distance from the pivot point to the center of mass of the child-swing system) decreases.
Analyze the change in moment of inertia: The moment of inertia \(I\) of the system also changes. Since the mass distribution moves closer to the pivot point (the suspension point of the swing), the moment of inertia \(I\) about the pivot point decreases. This is because moment of inertia depends on the mass and the square of its distance from the axis of rotation (\(I = \sum mr^2\)).
Determine the net effect on the time period: Both \(I\) and \(L\) decrease. However, the dominant effect is usually the decrease in \(L\). For a simple pendulum, the time period is given by \(T = 2\pi \sqrt{\frac{L}{g}}\). While a swing with a child is a physical pendulum, the effective length \(L\) (distance to the center of mass) is the key factor. When the child stands up, their center of mass moves closer to the pivot point, effectively reducing the length \(L\). Since \(T \propto \sqrt{L}\), a decrease in \(L\) will lead to a decrease in \(T\).
Conclusion: As the effective length (distance from the pivot to the center of mass) of the swing system decreases, the time period of the swing will decrease.
B) Decrease — When the child stands up, their center of mass moves closer to the pivot point of the swing. This effectively reduces the length of the pendulum (the distance from the pivot to the center of mass of the system). Since the time period of a pendulum is directly proportional to the square root of its effective length (\(T \propto \sqrt{L}\)), a decrease in effective length leads to a decrease in the time period.
The question asks us to identify the correct statements regarding oscillations. We need to evaluate each statement individually based on the fundamental principles of physics related to oscillations, including phase constant, resonance, free oscillations, and forced/damped oscillations.
In an oscillating system, the displacement \(x\) can often be described by an equation like \(x(t) = A \cos(\omega t + \phi)\), where \(A\) is the amplitude, \(\omega\) is the angular frequency, \(t\) is time, and \(\phi\) is the phase constant (or initial phase). The argument of the cosine function, \((\omega t + \phi)\), must be dimensionless because trigonometric functions operate on dimensionless quantities (angles are typically measured in radians, which are dimensionless). Since \(\omega t\) has units of radians (or is dimensionless if \(\omega\) is in \(\text{rad/s}\) and \(t\) in \(\text{s}\)), the phase constant \(\phi\) must also be dimensionless (in radians). Therefore, statement (i) is correct.
Resonance is a fundamental phenomenon in physics where a system oscillates with maximum amplitude at certain frequencies of excitation. This occurs when the frequency of the external driving force is equal or very close to the natural frequency of the oscillating system. At resonance, energy transfer from the driving force to the oscillator is most efficient, leading to a significant increase in the amplitude of oscillations. Therefore, statement (ii) is correct.
Free oscillations occur when a system is disturbed from its equilibrium position and then allowed to oscillate without any external periodic driving force. In an ideal scenario (without damping), such a system oscillates indefinitely at its characteristic frequency, known as its natural frequency (\(\omega\) or \(\nu\)). Examples include a simple pendulum swinging after being displaced or a mass on a spring oscillating after being stretched and released. Therefore, statement (iii) is correct.
In real-world scenarios, all oscillating systems experience some form of damping (e.g., air resistance, friction, internal resistance). These damping forces dissipate energy from the system, causing the amplitude of free oscillations to gradually decrease and eventually die out. To maintain oscillations at a constant amplitude, an external periodic force must be applied to continuously supply energy to the system, compensating for the energy lost due to damping. Such oscillations, maintained by an external driving force, are known as forced or driven oscillations. Therefore, statement (iv) is correct.
D) All the above — All four statements (i), (ii), (iii), and (iv) are correct descriptions of various aspects of oscillations.
The problem asks for the total distance covered by a particle undergoing Simple Harmonic Motion (SHM) in one complete time period. We need to understand the motion of a particle in SHM and how its position changes over one full oscillation.
Understanding SHM: In SHM, a particle oscillates back and forth about an equilibrium position. Let the equilibrium position be \(x=0\). The maximum displacement from the equilibrium position is called the amplitude, denoted by \(A\).
Motion in one time period: A complete oscillation (one time period, \(T\)) involves the particle starting from a point, moving through its path, and returning to the same point with the same velocity and direction of motion. Let's trace the path starting from the equilibrium position \(x=0\).
First quarter of the period (\(0\) to \(T/4\)): The particle moves from the equilibrium position (\(x=0\)) to one extreme position (\(x=+A\)). The distance covered is \(A\).
Second quarter of the period (\(T/4\) to \(T/2\)): The particle moves from the extreme position (\(x=+A\)) back to the equilibrium position (\(x=0\)). The distance covered is again \(A\).
Third quarter of the period (\(T/2\) to \(3T/4\)): The particle moves from the equilibrium position (\(x=0\)) to the other extreme position (\(x=-A\)). The distance covered is \(A\).
Fourth quarter of the period (\(3T/4\) to \(T\)): The particle moves from the extreme position (\(x=-A\)) back to the equilibrium position (\(x=0\)). The distance covered is again \(A\).
Total distance: Summing up the distances covered in each quarter of the period:
\[ \text{Total Distance} = A + A + A + A = 4A \]D) 4A — As explained above, the particle covers a distance of \(A\) from equilibrium to one extreme, another \(A\) from that extreme back to equilibrium, another \(A\) from equilibrium to the other extreme, and a final \(A\) from that extreme back to equilibrium. This sums up to \(4A\) in one complete time period.
The time period of a simple pendulum depends on the acceleration due to gravity. Inside an orbiting satellite, objects experience a state of apparent weightlessness, which means the effective acceleration due to gravity is zero. We will use the formula for the time period of a simple pendulum and substitute the effective gravity to find the result.
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.
A satellite orbiting the Earth is in a continuous state of free fall. Inside the satellite, everything, including the pendulum bob, experiences the same acceleration as the satellite itself. This creates a condition of apparent weightlessness.
In a state of apparent weightlessness, the effective acceleration due to gravity (\(g_{\text{eff}}\)) is zero. This is because the gravitational force acting on the object is entirely used to provide the centripetal force required for orbital motion, leading to no net force (or effective force) that would cause an object to fall relative to the satellite.
Substitute \(g_{\text{eff}} = 0\) into the time period formula:
\[ T = 2 \pi \sqrt{\frac{L}{0}} \]Division by zero results in an undefined value, which in the context of physics, means the time period is infinite. An infinite time period implies that the pendulum will not oscillate; it will simply float in its initial position, or if given a push, it will move in a straight line relative to the satellite's frame of reference without returning.
B) The time period becomes infinite because the effective acceleration due to gravity inside an orbiting satellite is zero. The pendulum will not oscillate.
This question involves the Doppler effect, which describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. When a source of sound moves away from a stationary observer, the observed frequency decreases.
D) It is halved. When the source moves away from a stationary observer, the observed frequency decreases. If the source moves away at the speed of sound, the frequency is exactly halved. This option represents a decrease in frequency, which is consistent with the Doppler effect for a receding source.
This question deals with the energy conservation in a simple pendulum. We need to understand the definitions of kinetic energy and potential energy in the context of a simple pendulum's motion, especially considering the reference point for potential energy.
If the reference level for potential energy is chosen such that the minimum potential energy (\(PE_{\text{min}}\)) is a positive value (e.g., the ground is below the lowest point of the pendulum's swing), then the maximum potential energy (\(PE_{\text{max}}\)) will be the sum of the maximum kinetic energy (\(E\)) and this minimum potential energy. Since \(PE_{\text{min}} > 0\), it implies that \(PE_{\text{max}} > E\).