Forced Oscillations and Resonance MCQ Quiz in मराठी - Objective Question with Answer for Forced Oscillations and Resonance - मोफत PDF डाउनलोड करा

CONCEPT:

Forced oscillations and resonance:

• When a system (such as a simple pendulum or a block attached to a spring) is displaced from its equilibrium position and released, it oscillates with its natural frequency ω, and the oscillations are called free oscillations.
• All free oscillations eventually die out because of the ever-present damping forces. However, an external agency can maintain these oscillations. These are called forced or driven oscillations.
• The most familiar example of forced oscillation is when a child in a garden swing periodically presses his feet against the ground (or someone else periodically gives the child a push) to maintain the oscillations.
• Suppose an external force F(t) of amplitude Fo that varies periodically with time is applied to a damped oscillator. Such a force can be represented as,

⇒ F(t) = Fo.cos(ωdt)

Where ωd = driving angular frequency

• When we apply the external periodic force to the oscillation, the oscillations with the natural frequency die out, and then the body oscillates with the angular frequency of the external periodic force. Its displacement at any time t, after the natural oscillations die out, is given as,

⇒ x(t) = A.cos(ωdt + ϕ)

Where A = amplitude and t is the time measured from the moment when we apply the periodic force

• The amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{[m^2(ω^2-ω_d^2)^2+ω_d^2b^2]^{1/2}}\)

a) Small damping (Driving frequency far from natural frequency):

• In this case, ωdb will be much smaller than \(m(ω^2-ω_d^2)\).
• We can neglect the term ωdb from the amplitude.
• The amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{m(ω^2-ω_d^2)}\)

• The amplitude of the oscillation is maximum when ω = ωd.

b) Driving frequency close to natural frequency:

• When ωd is very close to ω, \(m(ω^2-ω_d^2)\) would be much less than ωdb.
• For any reasonable value of b the amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{ω_db}\)

• This makes it clear that the maximum possible amplitude for a given driving frequency is governed by the driving frequency and the damping, and is never infinity.
• The phenomenon of increase in amplitude when the frequency of driving force is close to the natural frequency of the oscillator is called resonance.

EXPLANATION:

Given ω_P = 50 rad/sec, ωQ = 75 rad/sec, ωR = 80 rad/sec, ωS = 90 rad/sec, and ωd = 60 rad/sec

• We know that the amplitude of a simple pendulum for the forced oscillation is given as,

\(⇒ A=\frac{F_o}{[m^2(ω^2-ω_d^2)^2+ω_d^2b^2]^{1/2}}\)     -----(1)

Where ω = natural frequency, ωd = driving frequency, m = mass of the system, b = damping constant, and Fo = amplitude of the driving force

• By equation 1 it is clear that the amplitude of that pendulum will be the maximum for which the natural frequency is closest to the driving frequency.

⇒ ω_P -  ωd = -10

⇒ ω_Q -  ωd = 15

⇒ ω_R -  ωd = 20

⇒ ω_S -  ωd = 30

• Since the natural frequency of the pendulum P is closest to the driving frequency, so the amplitude of the pendulum P will be maximum. Hence, option 4 is correct.

Calculation:

Given:

First resonance column at depth, l₁ = 24 cm

Second resonance column at depth, l₂ = 78 cm

Third resonance column at depth, l₃ = ?

From the resonance condition:

l₁ + x = λ / 4 → (i)

l₂ + x = 3λ / 4 → (ii)

l₃ + x = 5λ / 4 → (iii)

From equations (i) and (ii), we have:

x = (l₂ - l₁) / 2 = (78 - 24) / 2 = 27 cm

Substituting this value of x into equation (iii), we get:

l₃ = 5(24) + 4(27) = 132 cm

∴ The third resonance is obtained at a depth of 132 cm.

Concept Used:

A charge q is placed at the midpoint of two fixed charges -q separated by a distance 2d. When displaced slightly by x perpendicular to the line joining the charges, it undergoes simple harmonic motion (SHM).

The force on the charge q is due to the electrostatic attraction from both -q charges.

Using Coulomb’s law, the net force acting on the displaced charge is proportional to x, indicating SHM.

The angular frequency ω for SHM is given by:

ω = √(2q² / 4πε₀d³m)

From the SHM formula, the time period T is given by:

T = 2π / ω

Calculation:

Given:

Charge on fixed particles = -q

Charge on moving particle = q

Mass of moving charge = m

Distance between fixed charges = 2d

The electrostatic force on charge q due to each -q is:

⇒ F = 2 × (q² / 4πε₀r²) × cos θ

Approximating for small displacements (x << d):

⇒ F ≈ (-2q² / 4πε₀d³) × x

Comparing with the standard SHM equation F = -kx, the effective force constant is:

⇒ k = 2q² / 4πε₀d³

For SHM, the angular frequency ω is:

⇒ ω = √(k / m)

⇒ ω = √[(2q² / 4πε₀d³m)]

Using T = 2π / ω:

⇒ T = 2π × √[(8π³ε₀m d³) / q²]

∴ The correct answer is: T = √(8ε₀ mπ² d / q²).

Calculation:

We are given the amplitude for a driven harmonic oscillator:

xm=Fm[ m2(ω2−ω02)2+b2ω2]12" id="MathJax-Element-291-Frame" role="presentation" style="position: relative;" tabindex="0">xm=Fm[ m2(ω2−ω20)2+b2ω2]12xm=Fm[m2(ω2−ω02)2+b2ω2]12 $\mathrm{x}_{\mathrm{m}}=\frac{\mathrm{F}_{\mathrm{m}}}{\left[\mathrm{~m}^{2}\left(\omega^{2}-\omega_{0}^{2}\right)^{2}+\mathrm{b}^{2} \omega^{2}\right]^{\frac{1}{2}}}$

where:

• Fm is the (constant) amplitude of the driving force,
• ω" id="MathJax-Element-292-Frame" role="presentation" style="position: relative;" tabindex="0">ωω $\omega$ is the frequency of the driving force,
• ω0" id="MathJax-Element-293-Frame" role="presentation" style="position: relative;" tabindex="0">ω0ω0 $\omega_0$ is the natural frequency of oscillation,
• m is the mass of the oscillator,
• b is the damping constant.

At resonance, the driving frequency ω" id="MathJax-Element-294-Frame" role="presentation" style="position: relative;" tabindex="0">ωω $\omega$ equals the natural frequency ω0" id="MathJax-Element-295-Frame" role="presentation" style="position: relative;" tabindex="0">ω0ω0 $\omega_0$ , i.e., ω=ω0" id="MathJax-Element-296-Frame" role="presentation" style="position: relative;" tabindex="0">ω=ω0ω=ω0 $\omega = \omega_0$ .

In this case, the amplitude formula simplifies as the term (ω2−ω02)" id="MathJax-Element-297-Frame" role="presentation" style="position: relative;" tabindex="0">(ω2−ω20)(ω2−ω02) $(\omega^2 - \omega_0^2)$ becomes zero:

xm=Fm[b2ω2]12=Fmbω" id="MathJax-Element-298-Frame" role="presentation" style="position: relative;" tabindex="0">xm=Fm[b2ω2]12=Fmbωxm=Fm[b2ω2]12=Fmbω $\mathrm{x}_{\mathrm{m}}=\frac{\mathrm{F}_{\mathrm{m}}}{\left[\mathrm{b}^{2} \omega^{2}\right]^{\frac{1}{2}}}=\frac{\mathrm{F}_{\mathrm{m}}}{\mathrm{b} \omega}$

The velocity amplitude, which is the maximum velocity, can be found by multiplying the amplitude by the angular frequency ω" id="MathJax-Element-299-Frame" role="presentation" style="position: relative;" tabindex="0">ωω $\omega$ :

Velocity Amplitude=ω⋅xm=ω⋅Fmbω=Fmb" id="MathJax-Element-300-Frame" role="presentation" style="position: relative;" tabindex="0">Velocity Amplitude=ω⋅xm=ω⋅Fmbω=FmbVelocity Amplitude=ω⋅xm=ω⋅Fmbω=Fmb $\text{Velocity Amplitude} = \omega \cdot \mathrm{x}_{\mathrm{m}} = \omega \cdot \frac{\mathrm{F}_{\mathrm{m}}}{\mathrm{b} \omega} = \frac{\mathrm{F}_{\mathrm{m}}}{\mathrm{b}}$

Final Answer: The velocity amplitude of the object at resonance is Fmb" id="MathJax-Element-301-Frame" role="presentation" style="position: relative;" tabindex="0">FmbFmb $\frac{F_{m}}{b}$ . Hence, the correct answer is option 3.

Concept Used:

The quality factor (Q) of a series LCR circuit is given by the ratio of the resonant frequency to the bandwidth of the circuit.

The formula for the quality factor Q in terms of angular frequency (ω) is:

Q = ω₀ / Δω

where, ω₀ is the resonant angular frequency and Δω is the bandwidth of the circuit.

Calculation:

Given data from the graph:

Resonant angular frequency, ω₀ = 1000 rad/s

Bandwith, Δω = ω_H - ω_L = 200 rad/s

Q = ω₀ / Δω

⇒ Q = 500 / 200

⇒ Q = 2.5

Conclusion:

∴ The quality factor (Q) of the series LCR circuit is 2.5.

Concept:

The resonance frequency\( (f_0) \)of an LCR circuit is given by:

\(f_0 = \frac{1}{2\pi\sqrt{L \cdot C}}\)

where L is the inductance and C is the capacitance. From the equation, it is evident that \(f_0\) is inversely proportional to the square root of the capacitance (C). As C increases, \(f_0 \)decreases.

Explanation:

• Option 1: The graph incorrectly shows a peak variation with C, which does not match the mathematical relationship of f₀ with C.
• Option 2: The graph incorrectly shows a linear increase of f₀ with C, which contradicts the inverse square root dependence.
• Option 3: The graph correctly represents an inverse relationship, where f₀ decreases with an increase in C, following the formula\( f_0 \propto \frac{1}{\sqrt{C}}\).
• Option 4: The graph incorrectlty shows the increasing of frequency with increasing capacitance.

Correct Answer: Option 3

The Correct answer is Resonance in ac circuits

Key Points

• Metal detectors often use a resonant LC (inductor-capacitor) circuit, which operates at a specific frequency.
• When a person carrying metal walks through the detector, the presence of the metal changes the inductance of the circuit, causing a shift in the resonant frequency.
• The detector is designed to recognize these shifts in resonance, and this deviation from the expected frequency triggers an alarm.

Here's how it works in detail:

• A metal detector generates an alternating current (AC) that creates an oscillating electromagnetic field.
• When metal is introduced, eddy currents are induced in the metal object, which alters the magnetic field and causes a change in the circuit's resonance.
• The detector circuit is set up to sense this change, and when it detects a deviation from the normal resonance, it raises an alarm.

CONCEPT:

• Damped oscillation: The oscillation of a body whose amplitude goes on decreasing with time is defined as a damped oscillation.
• Forced oscillation: The oscillation in which a body oscillates under the influence of an external periodic force is known as forced oscillation.
• The external agent which exerts the periodic force is called the driver and the oscillating system under consideration is called the driver body.

• Resonance: When the frequency of the vibrating body is equal to the natural frequency of another body, then the transfer of energy between the bodies is maximum and the other body vibrates with greater amplitude. This phenomenon is called resonance.

• So, under no damping condition, the resonance peak is sharp.

​EXPLANATION:

• Seeing by the above diagram, the damping force is small the resonance peak becomes sharp.

CONCEPT:

Forced oscillations and resonance:

• When a system (such as a simple pendulum or a block attached to a spring) is displaced from its equilibrium position and released, it oscillates with its natural frequency ω, and the oscillations are called free oscillations.
• All free oscillations eventually die out because of the ever-present damping forces. However, an external agency can maintain these oscillations. These are called forced or driven oscillations.
• The most familiar example of forced oscillation is when a child in a garden swing periodically presses his feet against the ground (or someone else periodically gives the child a push) to maintain the oscillations.
• Suppose an external force F(t) of amplitude Fo that varies periodically with time is applied to a damped oscillator. Such a force can be represented as,

⇒ F(t) = Fo.cos(ωdt)

Where ωd = driving angular frequency

• When we apply the external periodic force to the oscillation, the oscillations with the natural frequency die out, and then the body oscillates with the angular frequency of the external periodic force. Its displacement at any time t, after the natural oscillations die out, is given as,

⇒ x(t) = A.cos(ωdt + ϕ)

Where A = amplitude and t is the time measured from the moment when we apply the periodic force

• The amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{[m^2(ω^2-ω_d^2)^2+ω_d^2b^2]^{1/2}}\)

a) Small damping (Driving frequency far from natural frequency):

• In this case, ωdb will be much smaller than \(m(ω^2-ω_d^2)\).
• We can neglect the term ωdb from the amplitude.
• The amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{m(ω^2-ω_d^2)}\)

• The amplitude of the oscillation is maximum when ω = ωd.

b) Driving frequency close to natural frequency:

• When ωd is very close to ω, \(m(ω^2-ω_d^2)\) would be much less than ωdb.
• For any reasonable value of b the amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{ω_db}\)

• This makes it clear that the maximum possible amplitude for a given driving frequency is governed by the driving frequency and the damping, and is never infinity.
• The phenomenon of increase in amplitude when the frequency of driving force is close to the natural frequency of the oscillator is called resonance.

EXPLANATION:

• We know that for the forced oscillation of a damped spring-mass system, the amplitude of the system at resonance is given as,

\(⇒ A=\frac{F_o}{ω_db}\)

\(⇒ A\proptoω_d^{-1}\)

Where ωd = driving frequency, b = damping constant, and Fo = amplitude of the driving force

• Hence, option 2 is correct.

CONCEPT :

Forced oscillations and resonance:

• When a system (such as a simple pendulum or a block attached to a spring) is displaced from its equilibrium position and released, it oscillates with its natural frequency ω, and the oscillations are called free oscillations.
• All free oscillations eventually die out because of the ever-present damping forces. However, an external agency can maintain these oscillations. These are called forced or driven oscillations.
• The most familiar example of forced oscillation is when a child in a garden swing periodically presses his feet against the ground (or someone else periodically gives the child a push) to maintain the oscillations.
• Suppose an external force F(t) of amplitude Fo that varies periodically with time is applied to a damped oscillator. Such a force can be represented as,

⇒ F(t) = Fo.cos(ωdt)

Where ωd = driving angular frequency

• When we apply the external periodic force to the oscillation, the oscillations with the natural frequency die out, and then the body oscillates with the angular frequency of the external periodic force. Its displacement at any time t, after the natural oscillations die out, is given as,

⇒ x(t) = A.cos(ωdt + ϕ)

Where A = amplitude and t is the time measured from the moment when we apply the periodic force

• The amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{[m^2(ω^2-ω_d^2)^2+ω_d^2b^2]^{1/2}}\)

a) Small damping (Driving frequency far from natural frequency):

• In this case, ωdb will be much smaller than \(m(ω^2-ω_d^2)\).
• We can neglect the term ωdb from the amplitude.
• The amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{m(ω^2-ω_d^2)}\)

• The amplitude of the oscillation is maximum when ω = ωd.

b) Driving frequency close to natural frequency:

• When ωd is very close to ω, \(m(ω^2-ω_d^2)\) would be much less than ωdb.
• For any reasonable value of b the amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{ω_db}\)

• This makes it clear that the maximum possible amplitude for a given driving frequency is governed by the driving frequency and the damping, and is never infinity.
• The phenomenon of increase in amplitude when the frequency of driving force is close to the natural frequency of the oscillator is called resonance.

EXPLANATION:

Small damping (Driving frequency far from natural frequency):

• In this case, ωdb will be much smaller than \(m(ω^2-ω_d^2)\).
• We can neglect the term ωdb from the amplitude.
• The amplitude of the oscillations is given as,

\(⇒ A=\frac{F_o}{m(ω^2-ω_d^2)}\)

Where ω = natural frequency, ωd = driving frequency, m = mass of the system, and Fo = amplitude of the driving force

• Hence, option 2 is correct.