MIND GROW Magazine

📚 IIT–JEE Physics Complete Learning Hub

📌 Structured • Exam-Oriented • IIT–JEE Advanced Focused Learning

Stage-1 | Page-9 | IIT-JEE Advanced Level

Q65. A particle executes SHM with amplitude 10 cm and time period 2 s. Find the maximum velocity.

Solution:
ω = 2π / T = π rad/s
v_max = ωA = π × 0.10 = 0.314 m/s

Q66. At what displacement is the kinetic energy half of the maximum kinetic energy?

Solution:
KE = (1/2)KE_max
⇒ x = A / √2

Q67. Time taken by a particle to move from mean position to x = A/2 is?

Solution:
x = A sin(ωt)
sin(ωt) = 1/2 ⇒ ωt = π/6
t = T/12

Q68. Find the ratio of acceleration at x = A/2 to maximum acceleration.

Solution:
a = ω²x
a / a_max = (A/2)/A = 1/2

Q69. At what position does velocity become √3/2 of maximum velocity?

Solution:
v² = ω²(A² − x²)
x = A/2

Q70. A straight-line graph between acceleration and displacement is given. What does the slope represent?

Answer:
Slope = −ω²

Q71. Area under a velocity–time graph over one full cycle is:

Answer: Zero (net displacement)

Q72. Graph of potential energy vs displacement is:

Answer: Parabola opening upward

Q73. Phase difference between velocity and acceleration graphs is:

Answer: π/2

Q74. If slope of displacement–time graph is zero at a point, then that point is:

Answer: Extreme position

Q75. Ratio of time spent by particle in outer half (A/2 to A) to inner half (0 to A/2)?

Answer: More time in outer half (velocity smaller near extremes)

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