25 questions in 25 minutes on the NTA computer-based test interface, all from Oscillations. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Oscillations, of which 21 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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How many questions are in this Oscillations mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 21 of them are NEET previous-year questions.
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Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
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Physics
- Q1A particle is acted simultaneously by mutually perpendicular simple harmonic motions x = acosω t and y = asinω t. The trajectory of motion of the particle will be
- a parabola.
- a circle.
- an ellipse.
- a straight line.
Answer: (B) a circle.
C. Tests: eliminating time between two perpendicular SHMs to recover the path. Why C: square both and add, x² + y² = a²cos²ω t + a²sin²ω t = a². That is a circle of radius a about the origin, and since the phase advances uniformly the particle traces it at constant angular rate ω. Why not A: an ellipse (x²)/(a²) + (y²)/(b²) = 1 needs unequal amplitudes with the same π/2 phase gap, but here both amplitudes equal a, which closes the ellipse into a circle. Why not B: a parabola needs y ∝ x², which happens when one component oscillates at twice the frequency of the other, not at the same ω. Why not D: a straight line y = ± x comes from a phase difference of 0 or π, while here the phase difference is π/2 [misconception: ‘perpendicular SHMs always superpose along a line’]. Remember: equal amplitudes plus a quarter cycle phase gap traces a circle, which is uniform circular motion seen as two SHMs.
- Q2The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be:
- (T)/(√(2))
- √(2) T
- 2T
- T
Answer: (B) √(2) T
T = 2π√((M)/(k)). Doubling the mass to 2M: T’ = 2π√((2M)/(k)) = √(2) T.
- Q3Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. Reason (R): The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements, choose the correct answer from the options given below:
- Both (A) and (R) are true and (R) is the correct explanation of (A)
- (A) is false but (R) is true
- Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
- (A) is true but (R) is false
Answer: (C) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
D. Tests: computing surface g from GM/R² and then judging whether the offered reason is the operative one. Why D: on the planet gₚ = (G(4M))/((2R)²) = (4GM)/(4R²) = (GM)/(R²) = g_E. Since T = 2π√(l/g) and g is unchanged, the period is the same, so (A) is true. (R) is also true, the bob’s mass does not change on the trip, but it is not why (A) holds, because the period is mass independent in any case; the operative reason is gₚ = g_E. Why not A: (R) cannot be the explanation, because mass never enters T = 2π√(l/g), so an unchanged mass would not have saved the period had g changed [misconception: ‘any true reason printed under a true assertion must be its explanation’]. Why not B: (A) is true, since the factor 4 in mass and the factor 2² in radius cancel to give exactly the earth’s g. Why not C: (R) is a true statement in itself, nothing about moving the pendulum changes the mass of its bob. Remember: g ∝ M/R², so mass × 4 with radius × 2 leaves g, and therefore the period, untouched.
- Q4A particle executing linear S.H.M. has velocities V₁ and V₂ at distance x₁ and x₂ respectively, from the mean position, its angular velocity is
- √((V₂² – V₁²)/(x₁² – x₂²))
- √((V₂² – V₁²)/(x₂² – x₁²))
- √((V₁ V₂)/(x₁ + x₂))
- √((V₂² – V₁²)/(x₁ x₂))
Answer: (A) √((V₂² – V₁²)/(x₁² – x₂²))
A. Tests: eliminating the unknown amplitude between two applications of v² = ω²(A² – x²). Why A: V₁² = ω²(A² – x₁²) and V₂² = ω²(A² – x₂²). Subtracting the first from the second removes A²: V₂² – V₁² = ω²[(A² – x₂²) – (A² – x₁²)] = ω²(x₁² – x₂²), so ω = √((V₂² – V₁²)/(x₁² – x₂²)). Why not B: its denominator is subtracted the other way round, so numerator and denominator always carry opposite signs and the radicand is negative, which cannot give a real ω [misconception: ‘the order of subtraction does not matter under a root’]. Why not C: the elimination only ever produces x₁² – x₂², a product x₁x₂ never appears; it also fails the obvious check, since x₁ = x₂ forces V₁ = V₂ and this form then returns ω = 0 instead of leaving ω undetermined. Why not D: √(V₁V₂/(x₁ + x₂)) has units of √(m/s²), not s⁻¹, so it is not even dimensionally an angular frequency. Remember: two speeds at two positions kill the amplitude, so always subtract the two v² = ω²(A² – x²) equations.
- Q5A body is executing simple harmonic motion with frequency n. The frequency of its potential energy is:
- n
- 4n
- 2n
- 3n
Answer: (C) 2n
In SHM both kinetic and potential energy attain their maximum value twice in one complete oscillation (PE depends on x²). Hence the frequency of PE (and KE) variation is 2n, twice the oscillation frequency.
- Q6The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are:
- kg s⁻¹
- kg m s⁻²
- kg m s⁻¹
- kg s
Answer: (A) kg s⁻¹
F = kv ⇒ k = (F)/(v). Units = (kg m s⁻²)/(m s⁻¹) = kg s⁻¹.
- Q7When two displacements y₁ = asin(ω t) and y₂ = bcos(ω t) are superimposed, the resulting motion is:
- simple harmonic with amplitude (a)/(b)
- simple harmonic with amplitude (a+b)/(2)
- simple harmonic with amplitude √(a² + b²)
- not simple harmonic
Answer: (C) simple harmonic with amplitude √(a² + b²)
y₂ = bcosω t = bsin(ω t + (π)/(2)), so the two SHMs along the same line differ in phase by π/2. Their resultant is SHM with amplitude √(a² + b²).
- Q8Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:
- 11
- 10
- 8
- 9
Answer: (A) 11
A. Tests: turning T ∝ √(l) into a whole number oscillation count for a phase coincidence. Why A: (Tₛₕₒᵣₜ)/(T_(long)) = √((100)/(121)) = (10)/(11), so write Tₛₕₒᵣₜ = 10k and T_(long) = 11k. They are back in phase at the mean position when nₛ Tₛₕₒᵣₜ = nₗ T_(long), that is 10 nₛ = 11 nₗ. The smallest whole number solution is nₛ = 11 with nₗ = 10, so the shorter pendulum completes 11 vibrations. Why not B: 9 vibrations of the shorter pendulum last 90k, and 90k / 11k = 8.18 oscillations of the longer one, not a whole number. Why not C: 10 is the count belonging to the longer pendulum, not the shorter one, since 10 × 10k = 100k is only 9.09 oscillations of the longer pendulum [misconception: ‘the smaller number in the ratio belongs to the shorter pendulum’]. Why not D: 8 vibrations last 80k, which is 7.27 oscillations of the longer pendulum, so the two are not at the mean position together. Remember: the longer pendulum is the slower one, so the shorter one must fit in the larger count, 11 against 10.
- Q9A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10⁻² kg moving with a speed of 2 × 10² ms⁻¹. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g = 10 m/s²)
- 0.20 m
- 0.30 m
- 0.40 m
- 0.35 m
Answer: (A) 0.20 m
A. Tests: momentum conservation through a perfectly inelastic hit, followed by energy conservation on the swing. Why A: the bullet embeds, so the collision is inelastic and only momentum is conserved: (10⁻²)(2×10²) = 2 kg m/s = (1 + 0.01)v, giving v = 2/1.01 = 1.98 ms⁻¹. After the collision, energy is conserved as the bob swings up, (1)/(2)v² = gh, so h = (v²)/(2g) = (3.92)/(20) = 0.196 m, that is 0.20 m. Why not B: 0.40 m is exactly twice the answer, from writing h = v²/g and dropping the 2 in h = v²/2g [misconception: ‘the 2 in v² = 2gh is optional’]. Why not C: 0.30 m needs v = √(2 × 10 × 0.30) = 2.45 ms⁻¹, but momentum caps the bob speed at 1.98 ms⁻¹, so no consistent substitution gives it. Why not D: 0.35 m needs v = 2.65 ms⁻¹, even further above the 1.98 ms⁻¹ that momentum conservation allows. Remember: momentum through the collision, energy only after it, never energy through the embedding.
- Q10A pendulum clock keeping correct time on the ground is taken to the top of a high mountain. The clock will:
- Run fast
- Keep correct time
- Run slow
- Stop completely
Answer: (C) Run slow
At higher altitude g decreases, so T = 2π√(L/g) increases. A longer period means each ‘second’ takes more than a real second, so the clock runs slow (loses time).
- Q11When a damped harmonic oscillator completes 100 oscillations, its amplitude is reduced to (1)/(3) of its initial value. What will be its amplitude when it completes 200 oscillations?
- (2)/(3)
- (1)/(5)
- (1)/(6)
- (1)/(9)
Answer: (D) (1)/(9)
Amplitude decays exponentially: A = A₀ e⁻ᵇᵗ. After 100 oscillations e^(-b(100T)) = (1)/(3). After 200 oscillations the factor is e^(-b(200T)) = (e^(-b(100T)))² = ((1)/(3))² = (1)/(9) of the initial value.
- Q12The displacement of a particle executing SHM is y = A₀ + Asinω t + Bcosω t. The amplitude of its oscillation is:
- √(A₀² + (A+B)²)
- √(A² + B²)
- A₀ + √(A² + B²)
- A + B
Answer: (B) √(A² + B²)
A₀ only shifts the mean position (where y is constant), it does not contribute to amplitude. Combining Asinω t + Bcosω t = Rsin(ω t + φ) with R = √(A² + B²) (since the two terms are 90° out of phase). So amplitude = √(A² + B²).
- Q13In SHM the speed of the particle is maximum at the:
- Position where x = A/2
- Mean (equilibrium) position
- Extreme positions
- Position where acceleration is maximum
Answer: (B) Mean (equilibrium) position
v = ω√(A² – x²) is maximum when x = 0 (mean position), giving vₘₐₓ = Aω. At the extremes (x = ±A) the speed is zero.
- Q14Which condition must be satisfied for a body to execute simple harmonic motion?
- Acceleration is constant
- Acceleration is directly proportional to displacement and directed away from the mean position
- Acceleration is directly proportional to displacement and directed toward the mean position
- Velocity is proportional to displacement
Answer: (C) Acceleration is directly proportional to displacement and directed toward the mean position
SHM is defined by a restoring acceleration: a = -ω²x. The acceleration is proportional to displacement and always points back toward (not away from) the mean position, giving the negative sign.
- Q15A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. The pieces of ratio 2 and 3 are connected in series, giving force constant k’; if they are then connected in parallel the force constant is k”. The ratio k’ : k” is:
- 1 : 14
- 1 : 9
- 1 : 6
- 1 : 11
Answer: (D) 1 : 11
Cutting in ratio 1:2:3, the pieces have force constants 6k, 3k, 2k (constant ∝ 1/length, total length 6 parts). Take the 3k and 2k pieces. Series: (1)/(k’) = (1)/(3k) + (1)/(2k) = (5)/(6k) ⇒ k’ = (6k)/(5). Parallel: k” = 3k + 2k = 5k… Using all three as in the original (1:2:3 → 6k:3k:2k): series of all three (1)/(k’)=(1)/(6k)+(1)/(3k)+(1)/(2k)=(6)/(6k)⇒ k’=k; parallel k”=6k+3k+2k=11k. So k’:k” = 1:11.
- Q16Two oscillating simple pendulums with time periods T and (4T)/(3) are in phase at a given time. They will be again in phase after an elapse of time
- 4T
- 2T
- 5T
- 3T
Answer: (A) 4T
B. Tests: finding the smallest common multiple of two periods for a phase coincidence. Why B: the two are in phase again only when each has completed a whole number of oscillations, so n₁ T = n₂((4T)/(3)), which reduces to 3n₁ = 4n₂. The smallest whole numbers are n₁ = 4 and n₂ = 3, giving an elapsed time of 4T. The beat check agrees: (1)/(T) – (3)/(4T) = (1)/(4T), one coincidence every 4T. Why not A: 5T divided by (4T)/(3) is 3.75, not a whole number, so the slower pendulum is mid swing at that instant. Why not C: 3T divided by (4T)/(3) is 2.25, again not a whole number of oscillations for the slower pendulum. Why not D: 2T divided by (4T)/(3) is 1.5, so the slower pendulum has completed one and a half oscillations and is exactly out of phase [misconception: ‘half an extra oscillation still counts as in phase’]. Remember: they meet again at the LCM of the two periods, so count whole oscillations, not elapsed seconds.
- Q17If a simple harmonic oscillator has a displacement of 0.02 m and acceleration equal to 2.0 m/s² at any time, the angular frequency of the oscillator is equal to:
- 0.1 rad/s
- 1 rad/s
- 100 rad/s
- 10 rad/s
Answer: (D) 10 rad/s
|a| = ω² x ⇒ ω² = (a)/(x) = (2.0)/(0.02) = 100 ⇒ ω = 10 rad/s.
- Q18A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
- T/4
- T/8
- T/12
- T/2
Answer: (C) T/12
B. Tests: converting a phase angle into a fraction of the period from the mean position. Why B: measuring from the equilibrium position, x = asinω t. Half the amplitude means (a)/(2) = asinω t, so sinω t = (1)/(2) and the smallest solution is ω t = (π)/(6). With ω = (2π)/(T), t = (π)/(6)·(T)/(2π) = (T)/(12). Why not A: T/8 corresponds to ω t = (π)/(4), at which x = asin(π)/(4) = 0.707a, well past half the amplitude. Why not C: T/2 is half a full oscillation, which brings the particle back to the equilibrium position after having crossed (a)/(2) long before. Why not D: T/4 is the quarter period, the time to reach the full amplitude a, not half of it [misconception D: “half the displacement takes half the quarter period”]. Remember: from the mean the quarter period splits as T/12 to a/2, another T/12 to (√(3))/(2)a, and a final T/12 to a.
- Q19An object of mass 0.5 kg is executing simple harmonic motion. Its amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = (T)/(4) s starting from mean position. Assume that the initial phase of the oscillation is zero.
- 6.2 × 10⁻³ J
- 1.2 × 10³ J
- 6.2 × 10³ J
- 0.62 J
Answer: (D) 0.62 J
A. Tests: locating the particle at t = T/4 and evaluating U = (1)/(2)mω²x² in SI units. Why A: ω = (2π)/(T) = (2π)/(0.2) = 10π rad/s. Starting from the mean position with zero initial phase, x = Asinω t, and at t = (T)/(4) the phase is (π)/(2), so x = A = 0.05 m, the extreme position. Then U = (1)/(2)mω²x² = 0.5 × 0.5 × (10π)² × (0.05)² = 0.25 × 986.96 × 0.0025 = 0.617 J, about 0.62 J. Why not B: 6.2 × 10⁻³ J is a factor of 100 too small, which is what you get with x = 0.005 m, that is converting 5 cm as though it were 5 mm. Why not C: 1.2 × 10³ J does not follow from any consistent substitution; the two natural unit slips here give 6.2 × 10⁻³ J and 6.2 × 10³ J, and neither lands on 1.2. Why not D: 6.2 × 10³ J leaves the amplitude as the bare number 5 in metres: 0.25 × 986.96 × 25 = 6169 J [misconception D: “centimetres can be substituted directly into SI formulas”]. Remember: at t = T/4 from the mean the particle sits at full amplitude, so its PE there is the entire energy (1)/(2)mω²A².
- Q20Two springs of spring constants k₁ and k₂ are joined in series. The effective spring constant of the combination is:
- k₁ + k₂
- (k₁ k₂)/(k₁ + k₂)
- √(k₁ k₂)
- (k₁ + k₂)/(2)
Answer: (B) (k₁ k₂)/(k₁ + k₂)
For springs in series the same force stretches both, so extensions add: (1)/(k) = (1)/(k₁) + (1)/(k₂) ⇒ k = (k₁ k₂)/(k₁ + k₂). The series combination is softer than either spring.
- Q21A particle starts simple harmonic motion from the mean position. Its amplitude is a and time period is T. What is its displacement when its speed is half of its maximum speed?
- (√(2))/(3)a
- (√(3))/(2)a
- (a)/(√(2))
- (2)/(√(3))a
Answer: (B) (√(3))/(2)a
v = ω√(a² – x²), vₘₐₓ = ω a. Set ω√(a² – x²) = (1)/(2)ω a ⇒ a² – x² = (a²)/(4) ⇒ x² = (3a²)/(4) ⇒ x = (√(3))/(2)a.
- Q22Identify the function which represents a non-periodic motion.
- sin ω t
- sin(ω t + π/4)
- sin ω t + cos ω t
- e^(-ω t)
Answer: (D) e^(-ω t)
B. Tests: telling a repeating function apart from a monotonically decaying one. Why B: e^(-ω t) falls steadily from its initial value towards zero and never returns to a value it has already taken, so there is no T for which f(t + T) = f(t). It is a decay, not an oscillation. Why not A: sin(ω t + π/4) repeats every T = 2π/ω, the phase shift only moves the starting point of the cycle. Why not C: sinω t is the textbook periodic function, with T = 2π/ω. Why not D: sinω t + cosω t = √(2)sin(ω t + (π)/(4)), still periodic with T = 2π/ω [misconception: ‘adding two functions destroys periodicity’]. Remember: an exponential only ever decays, so anything of the form e^(-ω t) can never be periodic.
- Q23Two simple harmonic motions of angular frequency 100 and 1000 rad s⁻¹ have the same displacement amplitude. The ratio of their maximum acceleration is
- 1 : 10³
- 1 : 10²
- 1 : 10
- 1 : 10⁴
Answer: (B) 1 : 10²
D. Tests: the ω² dependence of maximum acceleration in SHM. Why D: aₘₐₓ = ω²A. With the same A, the ratio is (ω₁²)/(ω₂²) = ((100)/(1000))² = ((1)/(10))² = (1)/(100), that is 1 : 10². Why not A: 1 : 10³ cubes the frequency ratio, as if aₘₐₓ ∝ ω³. Why not B: 1 : 10⁴ raises the frequency ratio to the fourth power, which is the correct answer squared a second time. Why not C: 1 : 10 forgets to square at all; that is the ratio of maximum velocities, since vₘₐₓ = ω A [misconception C: “acceleration scales with ω like velocity does”]. Remember: vₘₐₓ carries one power of ω, aₘₐₓ carries two.
- Q24Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below:
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- (A) is false but (R) is true.
- (A) is true but (R) is false.
- Both (A) and (R) are true but (R) is not the correct explanation of (A).
Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
C. Tests: linking the fall of g with altitude to the pendulum period through T = 2π√(l/g). Why C: at height h above the surface, gₕ = gR²/(R+h)² < g, so g on a mountain top is smaller than at its base. Since T = 2π√(l/g) puts g in the denominator under the root, a smaller g gives a larger T, so (A) is true. (R) states exactly that inverse dependence, and it is the reason (A) holds, so (R) is true and it explains (A). Why not A: (R) is not an unrelated fact sitting beside (A), it is the very relation T ∝ 1/√(g) that makes the period longer up the mountain [misconception: ‘two true statements about one formula must be independent’]. Why not B: (R) is true, because g under a root in the denominator means T falls as g rises. Why not D: (A) is true, because g genuinely decreases with height above the earth’s surface. Remember: higher place, weaker g, slower pendulum, so a pendulum clock loses time on a mountain.
- Q25A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = (a)/(2) will be:
- (π a√(3))/(T)
- (π a√(3))/(2T)
- (π a)/(T)
- (3π² a)/(T)
Answer: (A) (π a√(3))/(T)
v = ω√(a² – x²) with ω = (2π)/(T). At x = (a)/(2): v = (2π)/(T)√(a² – (a²)/(4)) = (2π)/(T)·(a√(3))/(2) = (π a√(3))/(T).