Physics · Oscillations and Simple Harmonic Motion
Work through this past-paper style MCQ, then read the full explanation. Practice more physics questions on mMCQ with adaptive practice and topic analytics.
Two springs fixed at one end are stretched by 5 cm and 10 cm, respectively, when masses 0.5 kg and 1 kg are suspended at their lower ends. When displaced slightly from their mean positions and released, they will oscillate with time periods in the ratio
- A
1:√2
- B
1:2
- C
2:1
- D
√2:1
This is the correct answer. The time period of a spring-mass system is given by T = 2π√(m/k), where T is the time period, m is the mass, and k is the spring constant. Since the spring constant is the same for both springs (as the same force produces the same extension), the ratio of the time periods will only depend on the ratio of the masses: T1/T2 = √(m1/m2). Substituting the given masses, we get T1/T2 = √(0.5 kg / 1 kg) = √(1/2) = 1/√2.
This is the correct answer. The time period of a spring-mass system is given by T = 2π√(m/k), where T is the time period, m is the mass, and k is the spring constant. Since the spring constant is the same for both springs (as the same force produces the same extension), the ratio of the time periods will only depend on the ratio of the masses: T1/T2 = √(m1/m2). Substituting the given masses, we get T1/T2 = √(0.5 kg / 1 kg) = √(1/2) = 1/√2.
This is incorrect. The ratio of the time periods is not directly proportional to the ratio of the masses. Instead, it depends on the square root of the mass ratio.
This is incorrect. The ratio of the time periods is not inversely proportional to the ratio of the masses.
This is incorrect. The square root of the mass ratio is 1/√2, not √2.
Tagged under Physics · Oscillations and Simple Harmonic Motion · 2023