11th Physics MCQs Preparation 2022 and Onwards (Unit: Oscillation: Set-1 (27 MCQs)) for National MDCAT, ECAT, FPSC, PPSC, JOB Exam, Physics Lecturer, SSC, HSSC (F.Sc.), BS, MS Exams NTS, ETS ETc..
1. The acceleration of a body executing simple harmonic motion is
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2. What is the value of a spring constant when a 100 g mass is attached to a spring and it is accelerated 0.5 m s⁻² through a displacement of 5 cm?
F= Kx where F=ma so K = `\frac {ma}{x}` (All units must be converted to SI units before)
3. If a spring of forced constant K is cut into two equal parts, then the spring constant of each half is
F = Kx and F = K'\frac {x}{2}`, on comparing we get K' = 2k
4. When a body is performing S.H.M. then at its extreme position.
At extreme position, the body become at rest so velocity is zero
5. A particle is executing S.H.M along a straight line with amplitude A, its kinetic energy is maximum when its displacement is
Zero displacement means the body is at the mean position where speed is maximum so maximum K.E
T = 2π`\sqrt frac {m}{K}`
T = 2π`\sqrt frac {l}{g}`
T = `\frac {1}{f}` = `\frac {1}{2}` = 0.5 Hz
second's pendulum requires two seconds to complete one vibration.
P.E = `\frac {1}{2}`kx² and K.E = `\frac {1}{2}` k(x₀² - x²) so when P.E = K.E we get x = `\frac {x_0}{sqrt 2}` = 0.71 x₀
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Tuning of a radio set is an example of
Zero displacement means the body is at the mean position where speed is maximum so maximum K.E
6. The time period of a body attached to a spring depends upon.
T = 2π`\sqrt frac {m}{K}`
7. When the length of the pendulum is increased four times then its time period is increased. :
T = 2π`\sqrt frac {l}{g}`
8. What is the frequency of the body when its time period is 2 seconds?
T = `\frac {1}{f}` = `\frac {1}{2}` = 0.5 Hz
9. A second pendulum is one that has a time period of
second's pendulum requires two seconds to complete one vibration.
10. In S.H.M., at what distance from the mean position in terms of amplitude X₀, K.E., and P.E. both will have equal value?
P.E = `\frac {1}{2}`kx² and K.E = `\frac {1}{2}` k(x₀² - x²) so when P.E = K.E we get x = `\frac {x_0}{sqrt 2}` = 0.71 x₀
11. The instantaneous K.E of a mass attached to the end of an elastic spring is:
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12. In S.H.M., we have the conservation of
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13. A free oscillation has constant
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Under given conditions amplitude of the vibration is very much increased
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Given: f = `\ frac {1}{T}` = 1/2𝝿 `\sqrt frac {K}{m}` 👉 When f' = 2f then 👉 1/2𝝿`\sqrt frac {K}{m'}` = 2(1/2𝝿) `\sqrt frac {K}{m}` 👉 `\sqrt frac {K}{m'}` = 2 `\sqrt frac {K}{m}` 👉 taking square root both side and simplifying we get m' = `\ frac {1}{4}` m
Hook's Law 👉 Applied Force F ∝ displacement x
Hook's Law F = k₁x₁ = k₂x₂ 👉 k₁x₁ = k₂x₂ 👉 x₁ = `\ frac {k₂x₂}{k₁}` 👉 For both spring F = Kₑq (x₁ + x₂) 👉 Substituting value of x₁ 👉 F = Kₑq (`\ frac {k₂x₂}{k₁}` + x₂) 👉 substituting for F ( in term of spring 2) 👉 k₂x₂ = Kₑq (`\ frac {k₂x₂}{k₁}` + x₂) 👉 k₂ = Kₑq (`\ frac {k₂}{k₁}` + 1) Or 👉 Kₑq = k₁k₂/(k₁ + k₂)
21.
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14. When the damping force is equal to the oscillating force then the damping is called.
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15. A resonance occurs when the driving frequency is
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16.. In the absence of damping force, when the driving frequency is equal to the oscillating frequency then the amplitude of the oscillation will become
Under given conditions amplitude of the vibration is very much increased
17. Food is cooked in a microwave oven by the effect of: :
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18. To make the frequency double of a sprig oscillation, we have to:
Given: f = `\ frac {1}{T}` = 1/2𝝿 `\sqrt frac {K}{m}` 👉 When f' = 2f then 👉 1/2𝝿`\sqrt frac {K}{m'}` = 2(1/2𝝿) `\sqrt frac {K}{m}` 👉 `\sqrt frac {K}{m'}` = 2 `\sqrt frac {K}{m}` 👉 taking square root both side and simplifying we get m' = `\ frac {1}{4}` m
19. The restoring force of SHM is maximum when the particle:
Hook's Law 👉 Applied Force F ∝ displacement x
20. Two springs of spring constants k₁ and k₂ are joined in series. The effective spring constant of the combination is given by
Hook's Law F = k₁x₁ = k₂x₂ 👉 k₁x₁ = k₂x₂ 👉 x₁ = `\ frac {k₂x₂}{k₁}` 👉 For both spring F = Kₑq (x₁ + x₂) 👉 Substituting value of x₁ 👉 F = Kₑq (`\ frac {k₂x₂}{k₁}` + x₂) 👉 substituting for F ( in term of spring 2) 👉 k₂x₂ = Kₑq (`\ frac {k₂x₂}{k₁}` + x₂) 👉 k₂ = Kₑq (`\ frac {k₂}{k₁}` + 1) Or 👉 Kₑq = k₁k₂/(k₁ + k₂)
21.
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22. The heating and cooking of food evenly by a microwave oven is an example of:
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23. The time period of the same pendulum at Karachi and Murree is related as
The equation for Time Period T = 2𝝿 `\sqrt frac {L}{g}` 👉 shows that Time Period is inversely proportional to the square root of g 👉 Value of g is slightly greater at Karachi (Elevation = 10 m) than at Murree (Elevation = = 2300 m)(as the value of g depends on height) 👉 Keeping the same pendulum (no change in length) Tₖ will be less than Tₘ
24. In an isolated system the total energy of the vibrating mass and spring is:
Law of conservation of energy
25. While deriving the equation of time period for a simple pendulum which quantity should be kept small:
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26. If the period of oscillation of mass (M) suspended from a spring is 2s, then the period of mass 4M will be
Equation for Time Period of mass attached to spring T = 2𝝿 `\sqrt frac {M}{K}` = 2 s 👉 When M' = 4M then 👉 T' = 2𝝿 `\sqrt frac {4M}{K}` = 2𝝿 x 2 `\sqrt frac {M}{K}` = 2 (2𝝿 `\sqrt frac {M}{K}`) = 2 (2 s) = 4 s
27. The time period of a simple pendulum is 2 seconds. If its length is increased by 4 times then its period becomes :
Equation for Time Period of mass attached to spring T = 2𝝿 `\sqrt frac {L}{g}` = 2 s 👉 When L' = 4L then 👉 T' = 2𝝿 `\sqrt frac {4L}{g}` = 2𝝿 x 2 `\sqrt frac {L}{g}` = 2 (2𝝿 `\sqrt frac {L}{g}`) = 2 (2 s) = 4 s
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