HEC USAT-E (Pre-Engineering) Physics: Oscillations MCQs

HEC USAT-E (Pre-Engineering) Physics: Oscillations MCQs — sample questions

1. Question 1

   Q1. A particle executes SHM with amplitude 4 cm. At what displacement is its energy half kinetic and half potential?

   • A) 2 cm
   • B) 2 √2 cm
   • C) 4 / √2 cm
   • D) 1 cm

   Answer: 2 √2 cm

   Explanation: In SHM, energy is half kinetic and half potential when displacement is A / √2. Here, A = 4 cm, so displacement = 4 / √2 = 2 √2 cm.

2. Question 2

   Q2. A simple pendulum has a time period of 2 s. If its length is increased by 4 times, its new time period is

   • A) 4 s
   • B) 2 s
   • C) 8 s
   • D) 1 s

   Answer: 4 s

   Explanation: Time period of simple pendulum, T = 2 π √(l / g). When l is increased 4 times, T becomes 2 times, so new T = 2 * 2 = 4 s.

3. Question 3

   Q3. The equation of motion of a particle is d²x / dt² + kx = 0. The frequency of vibration is

   • A) 1 / 2 π √(k)
   • B) 2 π √k
   • C) √(k) / 2 π
   • D) 1 / 2 π √k

   Answer: 1 / 2 π √(k)

   Explanation: The given equation represents SHM with angular frequency ω = √k. Frequency, f = ω / 2 π = 1 / 2 π √k.

4. Question 4

   Q4. A particle is executing SHM with maximum velocity 10 cm/s and maximum acceleration 40 cm/s². Its amplitude is

   • A) 2.5 cm
   • B) 5 cm
   • C) 10 cm
   • D) 1.25 cm

   Answer: 2.5 cm

   Explanation: In SHM, v_max = Aω and a_max = Aω². Given v_max = 10 and a_max = 40. Dividing, ω = 4. So, A = v_max / ω = 10 / 4 = 2.5 cm.

5. Question 5

   Q5. The time period of a simple pendulum is 1 s. If its length is halved, its new time period is

   • A) 1 / √2 s
   • B) √2 s
   • C) 1 s
   • D) 2 s

   Answer: 1 / √2 s

   Explanation: T ∝ √l. When l is halved, T becomes 1 / √2 times, so new T = 1 / √2 s.

6. Question 6

   Q6. A simple harmonic oscillator has amplitude A and time period T. Its maximum speed is

   • A) A / T
   • B) 2 π A / T
   • C) A / 2 T
   • D) π A / T

   Answer: 2 π A / T

   Explanation: Maximum speed = Aω = A * 2 π / T = 2 π A / T.

7. Question 7

   Q7. The potential energy of a simple harmonic oscillator is maximum at

   • A) mean position
   • B) extreme position
   • C) between mean and extreme position
   • D) any position

   Answer: extreme position

   Explanation: In SHM, potential energy is maximum at extreme positions where displacement is maximum.

8. Question 8

   Q8. A particle executes SHM with frequency f. The frequency of its kinetic energy is

   • A) f
   • B) 2 f
   • C) f / 2
   • D) 4 f

   Answer: 2 f

   Explanation: Kinetic energy oscillates at twice the frequency of displacement, so its frequency is 2 f.

9. Question 9

   Q9. The displacement of a simple harmonic oscillator is given by x = A sin(ωt + φ). Its velocity is

   • A) Aω cos(ωt + φ)
   • B) -Aω sin(ωt + φ)
   • C) Aω² sin(ωt + φ)
   • D) -Aω² cos(ωt + φ)

   Answer: Aω cos(ωt + φ)

   Explanation: Velocity is the derivative of displacement. So, v = d(A sin(ωt + φ))/dt = Aω cos(ωt + φ).

10. Question 10

    Q10. The total energy of a simple harmonic oscillator is proportional to

    • A) amplitude
    • B) square of amplitude
    • C) frequency
    • D) square of frequency

    Answer: square of amplitude

    Explanation: Total energy of SHM = 1/2 mω²A², which is proportional to A², the square of amplitude.

11. Question 11

    Q11. A simple pendulum is taken to a height equal to earth's radius. Its time period becomes

    • A) 2 T
    • B) √2 T
    • C) T / 2
    • D) T / √2

    Answer: 2 T

    Explanation: At height R, g' = g / 4. T ∝ 1 / √g, so T' = 2 T.

12. Question 12

    Q12. The equation of a simple harmonic motion is given by y = 6 sin(10t) + 8 cos(10t). Its amplitude is

    • A) 10
    • B) 14
    • C) 2
    • D) 5

    Answer: 10

    Explanation: The given equation can be written as y = A sin(10t + φ). Using trigonometric identity, A = √(6² + 8²) = √(36 + 64) = √100 = 10.

13. Question 13

    Q13. A particle is executing SHM. The phase difference between its velocity and displacement is

    • A) π / 2
    • B) π
    • C) 0
    • D) 3π / 2

    Answer: π / 2

    Explanation: In SHM, velocity leads displacement by π / 2.

14. Question 14

    Q14. The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, its new time period is

    • A) T / 2
    • B) T / 4
    • C) 2 T
    • D) 4 T

    Answer: T / 2

    Explanation: T ∝ 1 / √k. When spring is cut into 4 parts, k becomes 4 times, so T becomes 1 / 2.

15. Question 15

    Q15. The displacement of a particle executing SHM is given by y = A sin(ωt). The maximum acceleration is

    • A) Aω
    • B) Aω²
    • C) -Aω²
    • D) A / ω

    Answer: Aω²

    Explanation: Maximum acceleration = Aω², which occurs at extreme positions.

16. Question 16

    Q16. The ratio of maximum acceleration to maximum velocity in SHM is

    • A) ω
    • B) 1 / ω
    • C) ω²
    • D) 1 / ω²

    Answer: ω

    Explanation: Maximum acceleration = Aω² and maximum velocity = Aω. Their ratio = ω.

17. Question 17

    Q17. A simple pendulum is oscillating with angular frequency ω. Its time period is

    • A) 2 π / ω
    • B) ω / 2 π
    • C) 1 / ω
    • D) ω

    Answer: 2 π / ω

    Explanation: Time period = 2 π / ω by definition of angular frequency.

18. Question 18

    Q18. The kinetic energy of a simple harmonic oscillator is maximum at

    • A) mean position
    • B) extreme position
    • C) any position
    • D) between mean and extreme

    Answer: mean position

    Explanation: In SHM, kinetic energy is maximum at mean position where velocity is maximum.

19. Question 19

    Q19. A particle executes SHM with amplitude A. The distance traveled by it in one time period is

    • A) A
    • B) 2 A
    • C) 4 A
    • D) 8 A

    Answer: 4 A

    Explanation: In one time period, the particle travels from one extreme to other and back, so total distance = 4 A.

20. Question 20

    Q20. The frequency of a simple harmonic oscillator is f. If its amplitude is doubled, its new frequency is

    • A) f
    • B) 2 f
    • C) f / 2
    • D) 4 f

    Answer: f

    Explanation: Frequency of SHM is independent of amplitude, so it remains the same.