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Simple Harmonic Motion (SHM) – Tough Objective Questions
Stage-1 | Page-6 | Numerical Traps & Mixed Concepts
Q36. A particle executes SHM with amplitude 10 cm and time period 2 s.
Find the maximum speed.
Options:
A) 10π cm/s B) 20π cm/s C) 5π cm/s D) 40π cm/s
Options:
A) 10π cm/s B) 20π cm/s C) 5π cm/s D) 40π cm/s
Answer: B
Solution:
ω = 2π/T = π rad/s vmax = ωA = π × 20 = 20π cm/s
Solution:
ω = 2π/T = π rad/s vmax = ωA = π × 20 = 20π cm/s
Q37. At what position is the acceleration half of maximum acceleration?
Options:
A) x = A/2 B) x = A/√2 C) x = A/4 D) x = A
Options:
A) x = A/2 B) x = A/√2 C) x = A/4 D) x = A
Answer: A
Explanation:
a = ω²x a = (1/2)amax ⇒ x = A/2
Explanation:
a = ω²x a = (1/2)amax ⇒ x = A/2
Q38. If the total energy of SHM is E, potential energy at x = A/√2 is:
Options:
A) E/4 B) E/2 C) 3E/4 D) E
Options:
A) E/4 B) E/2 C) 3E/4 D) E
Answer: B
Calculation:
PE = E(x²/A²) = E(1/2)
Calculation:
PE = E(x²/A²) = E(1/2)
Q39. Ratio of speeds of a particle at x = A/2 and x = A/√2 is:
Options:
A) √3 : 1 B) 1 : √3 C) 2 : 1 D) 1 : 2
Options:
A) √3 : 1 B) 1 : √3 C) 2 : 1 D) 1 : 2
Answer: A
Explanation:
v = ω√(A² − x²) At A/2 → √(3A²/4) At A/√2 → √(A²/2)
Explanation:
v = ω√(A² − x²) At A/2 → √(3A²/4) At A/√2 → √(A²/2)
Q40. The phase difference between velocity and acceleration in SHM is:
Options:
A) 0 B) π/2 C) π D) 3π/2
Options:
A) 0 B) π/2 C) π D) 3π/2
Answer: B
Q41. A particle covers distances x and y in equal successive intervals of time from mean position.
Then:
Options:
A) x = y B) x > y C) x < y D) Data insufficient
Options:
A) x = y B) x > y C) x < y D) Data insufficient
Answer: B
Concept:
Speed decreases as particle moves away from mean.
Concept:
Speed decreases as particle moves away from mean.
Q42. In SHM, which quantity remains constant?
Options:
A) Velocity B) Acceleration C) Total Energy D) Momentum
Options:
A) Velocity B) Acceleration C) Total Energy D) Momentum
Answer: C
Q43. The restoring force in SHM is maximum when:
Options:
A) x = 0 B) x = A/2 C) x = A D) Always constant
Options:
A) x = 0 B) x = A/2 C) x = A D) Always constant
Answer: C
Q44. The ratio of maximum acceleration to maximum velocity is:
Options:
A) ω B) 1/ω C) ω² D) A
Options:
A) ω B) 1/ω C) ω² D) A
Answer: A
Reason:
amax/vmax = ω²A / (ωA) = ω
Reason:
amax/vmax = ω²A / (ωA) = ω
Q45. The average velocity of a particle in SHM over half cycle is:
Options:
A) Zero B) 2A/T C) 4A/T D) ωA
Options:
A) Zero B) 2A/T C) 4A/T D) ωA
Answer: B
Explanation:
Average velocity = displacement / time = 2A / (T/2)
Explanation:
Average velocity = displacement / time = 2A / (T/2)
✅ Stage-1 | Page-6 Completed Successfully
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