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A straight conducting bar of mass m and length (is suspended horizontally with two non-conducting 3. The negativel springs of stiffness k as shown in figure The capacitor is initially charged to the potential difference V_{1} At time t = 0 the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane. The amplitude of these oscillations (assuming the time of discharge of capacitor is much smaller than the Period T of the mechanical oscillation of the bar) is:?
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A straight conducting bar of mass m and length (is suspended horizonta...
The given system consists of a conducting bar, two non-conducting springs, and a capacitor. When the switch S is closed, the capacitor discharges and the bar starts oscillating in a vertical plane.
**Analysis of the System:**
1. **Initial State:**
- The capacitor is charged to a potential difference V₁. This means that one plate of the capacitor has a charge +Q and the other plate has a charge -Q.
- The bar is initially in a horizontal position and is at rest.
- The springs are in their natural lengths and exert no force on the bar.
2. **Switch Closure:**
- When the switch S is closed, the capacitor discharges and the charges on its plates start decreasing.
- As the charges on the plates change, an electric current is induced in the conducting bar due to electromagnetic induction. This current produces a magnetic field around the bar.
- According to Ampere's law, the magnetic field exerts a force on the current-carrying bar in a direction perpendicular to both the current and the magnetic field. This force causes the bar to move.
3. **Mechanical Oscillation:**
- As the bar moves due to the magnetic force, it stretches one spring and compresses the other, resulting in a restoring force.
- The restoring force acts to bring the bar back to its equilibrium position, causing it to oscillate in a vertical plane.
- The period of oscillation (T) depends on the stiffness of the springs, the mass of the bar, and the length of the bar.
**Amplitude of Oscillations:**
The amplitude of the oscillations can be determined by considering the conservation of energy in the system. When the bar is at its maximum displacement, all the initial electrical energy stored in the capacitor is converted into potential energy of the bar and the springs.
The potential energy of the bar-spring system is given by:
U = 1/2 k x₁² + 1/2 k x₂²
where k is the stiffness of the springs, x₁ is the displacement of the stretched spring, and x₂ is the displacement of the compressed spring.
The initial electrical energy stored in the capacitor is given by:
E = 1/2 C V₁²
where C is the capacitance of the capacitor.
Since the total energy in the system is conserved, we have:
U + E = constant
At maximum displacement, the potential energy is maximum and the electrical energy is zero. Therefore:
1/2 k x₁² + 1/2 k x₂² = 0
This implies that x₁ = -x₂, i.e., the displacements of the two springs are equal in magnitude but opposite in direction.
Thus, the amplitude of the oscillations is given by:
A = |x₁| = |x₂|
Therefore, the amplitude of the oscillations is equal to the displacement of either spring, which depends on the initial charge on the capacitor and the stiffness of the springs.
**Analysis of the System:**
1. **Initial State:**
- The capacitor is charged to a potential difference V₁. This means that one plate of the capacitor has a charge +Q and the other plate has a charge -Q.
- The bar is initially in a horizontal position and is at rest.
- The springs are in their natural lengths and exert no force on the bar.
2. **Switch Closure:**
- When the switch S is closed, the capacitor discharges and the charges on its plates start decreasing.
- As the charges on the plates change, an electric current is induced in the conducting bar due to electromagnetic induction. This current produces a magnetic field around the bar.
- According to Ampere's law, the magnetic field exerts a force on the current-carrying bar in a direction perpendicular to both the current and the magnetic field. This force causes the bar to move.
3. **Mechanical Oscillation:**
- As the bar moves due to the magnetic force, it stretches one spring and compresses the other, resulting in a restoring force.
- The restoring force acts to bring the bar back to its equilibrium position, causing it to oscillate in a vertical plane.
- The period of oscillation (T) depends on the stiffness of the springs, the mass of the bar, and the length of the bar.
**Amplitude of Oscillations:**
The amplitude of the oscillations can be determined by considering the conservation of energy in the system. When the bar is at its maximum displacement, all the initial electrical energy stored in the capacitor is converted into potential energy of the bar and the springs.
The potential energy of the bar-spring system is given by:
U = 1/2 k x₁² + 1/2 k x₂²
where k is the stiffness of the springs, x₁ is the displacement of the stretched spring, and x₂ is the displacement of the compressed spring.
The initial electrical energy stored in the capacitor is given by:
E = 1/2 C V₁²
where C is the capacitance of the capacitor.
Since the total energy in the system is conserved, we have:
U + E = constant
At maximum displacement, the potential energy is maximum and the electrical energy is zero. Therefore:
1/2 k x₁² + 1/2 k x₂² = 0
This implies that x₁ = -x₂, i.e., the displacements of the two springs are equal in magnitude but opposite in direction.
Thus, the amplitude of the oscillations is given by:
A = |x₁| = |x₂|
Therefore, the amplitude of the oscillations is equal to the displacement of either spring, which depends on the initial charge on the capacitor and the stiffness of the springs.
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A straight conducting bar of mass m and length (is suspended horizontally with two non-conducting 3. The negativel springs of stiffness k as shown in figure The capacitor is initially charged to the potential difference V_{1} At time t = 0 the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane. The amplitude of these oscillations (assuming the time of discharge of capacitor is much smaller than the Period T of the mechanical oscillation of the bar) is:?
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A straight conducting bar of mass m and length (is suspended horizontally with two non-conducting 3. The negativel springs of stiffness k as shown in figure The capacitor is initially charged to the potential difference V_{1} At time t = 0 the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane. The amplitude of these oscillations (assuming the time of discharge of capacitor is much smaller than the Period T of the mechanical oscillation of the bar) is:? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A straight conducting bar of mass m and length (is suspended horizontally with two non-conducting 3. The negativel springs of stiffness k as shown in figure The capacitor is initially charged to the potential difference V_{1} At time t = 0 the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane. The amplitude of these oscillations (assuming the time of discharge of capacitor is much smaller than the Period T of the mechanical oscillation of the bar) is:? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A straight conducting bar of mass m and length (is suspended horizontally with two non-conducting 3. The negativel springs of stiffness k as shown in figure The capacitor is initially charged to the potential difference V_{1} At time t = 0 the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane. The amplitude of these oscillations (assuming the time of discharge of capacitor is much smaller than the Period T of the mechanical oscillation of the bar) is:?.
A straight conducting bar of mass m and length (is suspended horizontally with two non-conducting 3. The negativel springs of stiffness k as shown in figure The capacitor is initially charged to the potential difference V_{1} At time t = 0 the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane. The amplitude of these oscillations (assuming the time of discharge of capacitor is much smaller than the Period T of the mechanical oscillation of the bar) is:? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A straight conducting bar of mass m and length (is suspended horizontally with two non-conducting 3. The negativel springs of stiffness k as shown in figure The capacitor is initially charged to the potential difference V_{1} At time t = 0 the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane. The amplitude of these oscillations (assuming the time of discharge of capacitor is much smaller than the Period T of the mechanical oscillation of the bar) is:? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A straight conducting bar of mass m and length (is suspended horizontally with two non-conducting 3. The negativel springs of stiffness k as shown in figure The capacitor is initially charged to the potential difference V_{1} At time t = 0 the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane. The amplitude of these oscillations (assuming the time of discharge of capacitor is much smaller than the Period T of the mechanical oscillation of the bar) is:?.
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