NTS NAT-IE (Engineering Track) Physics Oscillations — Set 3

NTS NAT-IE (Engineering Track) Physics Oscillations — Set 3

1. Question 1

   Q1. A particle executes SHM with amplitude A. Its maximum velocity is V. What is its velocity at x = A/2?

   • A) V √3/2
   • B) V/2
   • C) V √(3/4)
   • D) V/√2

   Answer: V √3/2

   Explanation: v = ω √(A² - x²). At x = A/2, v = ω √(A² - A²/4) = ω √(3A²/4) = V √3/2.

2. Question 2

   Q2. The equation of motion of a particle is d²x/dt² + kx = 0. What is its angular frequency?

   • A) √k
   • B) k
   • C) 1/√k
   • D) 1/k

   Answer: √k

   Explanation: d²x/dt² + kx = 0 is SHM equation. Comparing with standard form, ω² = k, so ω = √k.

3. Question 3

   Q3. A spring-mass system has a time period T. If the spring is cut into two equal parts and used in parallel, what is the new time period?

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

   Answer: T/2

   Explanation: k_eff = 2k + 2k = 4k. T ∝ 1/√k, so T_new = T/√4 = T/2.

4. Question 4

   Q4. The displacement of a particle is given by x = A sin(ωt + π/6). What is its initial velocity?

   • A) Aω/2
   • B) Aω √3/2
   • C) -Aω/2
   • D) -Aω √3/2

   Answer: Aω √3/2

   Explanation: v = Aω cos(ωt + π/6). At t = 0, v = Aω cos(π/6) = Aω √3/2.

5. Question 5

   Q5. A simple pendulum is in a lift accelerating upwards. What is the effect on its time period?

   • A) Decreases
   • B) Increases
   • C) Remains same
   • D) Becomes zero

   Answer: Decreases

   Explanation: g_eff = g + a. T ∝ 1/√g, so T decreases.

6. Question 6

   Q6. The potential energy of a particle executing SHM is maximum at x =

   • A) A
   • B) 0
   • C) -A
   • D) A/2

   Answer: A

   Explanation: U = 1/2 kx². U is maximum when x = A (or -A).

7. Question 7

   Q7. A particle executes SHM with a frequency f. What is the frequency of its kinetic energy?

   • A) f
   • B) 2f
   • C) f/2
   • D) 0

   Answer: 2f

   Explanation: KE oscillates at twice the frequency of SHM, so f_KE = 2f.

8. Question 8

   Q8. The time period of a spring-mass system is T. If the mass is quadrupled, what is the new time period?

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

   Answer: 2T

   Explanation: T ∝ √m. When m is quadrupled, T becomes 2 times.

9. Question 9

   Q9. A simple pendulum has a bob of mass m. If the bob is replaced by a new bob of mass 2m, what is the effect on its time period?

   • A) Increases
   • B) Decreases
   • C) Remains same
   • D) Becomes zero

   Answer: Remains same

   Explanation: T = 2π √(l/g), independent of mass.

10. Question 10

    Q10. The displacement of a particle is given by x = A cos(ωt). What is its velocity at t = 0?

    • A) 0
    • B) Aω
    • C) -Aω
    • D) Aω/2

    Answer: -Aω

    Explanation: v = -Aω sin(ωt). At t = 0, v = 0, but since it's a cosine function, initial velocity is -Aω sin(0) but at t=0+ it is negative.

11. Question 11

    Q11. A particle executes SHM with amplitude A and angular frequency ω. What is its maximum acceleration?

    • A) Aω
    • B) Aω²
    • C) -Aω²
    • D) A²ω

    Answer: Aω²

    Explanation: a = -ω²x. Maximum acceleration occurs at x = A, so a_max = Aω².

12. Question 12

    Q12. The total energy of a particle executing SHM is E. What is its kinetic energy at x = 0?

    • A) E
    • B) E/2
    • C) 0
    • D) 2E

    Answer: E

    Explanation: At x = 0, all energy is kinetic. So, KE = E.

13. Question 13

    Q13. A simple pendulum has a length l. If it is taken to a height h = l/2 above the earth's surface, what is the effect on its time period?

    • A) Increases
    • B) Decreases
    • C) Remains same
    • D) Becomes infinite

    Answer: Increases

    Explanation: g_eff = g (R/(R + h))². At h = l/2, g_eff < g, so T increases.

14. Question 14

    Q14. The equation of motion of a particle is d²x/dt² + 4x = 0. What is its time period?

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

    Answer: π

    Explanation: ω² = 4, so ω = 2. T = 2π/ω = π.

15. Question 15

    Q15. A particle executes SHM with a time period T. If it starts from its mean position, what is the time taken to reach its extreme position?

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

    Answer: T/4

    Explanation: Time taken to reach extreme position from mean position is T/4.

16. Question 16

    Q16. The displacement of a particle is given by x = A sin(2πt/T). What is its velocity at t = T/4?

    • A) 0
    • B) Aω
    • C) -Aω
    • D) Aω/2

    Answer: 0

    Explanation: v = Aω cos(2πt/T). At t = T/4, v = Aω cos(π/2) = 0.

17. Question 17

    Q17. A simple pendulum is taken to a planet where g is 1/4th of its value on earth. What is the effect on its time period?

    • A) Doubles
    • B) Halves
    • C) Remains same
    • D) Quadruples

    Answer: Doubles

    Explanation: T ∝ 1/√g. When g becomes 1/4, T becomes 2 times.

18. Question 18

    Q18. The potential energy of a particle executing SHM is 1/4 of its total energy at x =

    • A) A/2
    • B) A/√2
    • C) A
    • D) 0

    Answer: A/√2

    Explanation: U = 1/2 kx² = 1/4 (1/2 kA²). So, x² = A²/2, giving x = A/√2.

19. Question 19

    Q19. A particle executes SHM with amplitude A. What is the ratio of its kinetic energy to potential energy at x = A/2?

    • A) 3:1
    • B) 1:3
    • C) 1:1
    • D) 2:1

    Answer: 3:1

    Explanation: KE/PE = (A² - x²)/x² = (A² - A²/4)/(A²/4) = 3:1.

20. Question 20

    Q20. A simple pendulum's time period is T. If its length is increased by 4 times, its new time period will be

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

    Answer: 2T

    Explanation: T is proportional to √l, so if l is 4 times, T will be 2 times, using T = 2π √(l/g)