Oscillations and Waves MCQs with Answers – Part 1 (NEET Physics)

Explanation: Oscillatory motion involves repeated movement to and fro about an equilibrium or mean position. The moving body passes through its equilibrium position in opposite directions during successive parts of the motion. It also reverses its direction at the turning points of its path. A block moving repeatedly on either side of a stable equilibrium position exhibits this characteristic behaviour. Uniform circular motion is periodic because the moving point returns to the same position and velocity after each revolution. However, the point itself travels around a closed circular path rather than moving to and fro along a fixed line about an equilibrium position. Similarly, a satellite in an ideal circular orbit undergoes periodic revolution rather than the specified oscillatory motion. Thus, repetition after a definite time interval does not by itself establish that a motion is oscillatory. The essential distinction is the repeated reversal of motion about an equilibrium position.

Explanation: \(\textbf{Given:}\) The extreme coordinates of the symmetric oscillation are \[ x_{\min}=-9\,\text{cm},\qquad x_{\max}=+3\,\text{cm} \] \(\textbf{Equilibrium position:}\) For symmetric oscillation, the equilibrium position lies midway between the two extreme positions. \[ x_{\mathrm{eq}}=\frac{x_{\max}+x_{\min}}{2} \] Substituting the coordinates, \[ x_{\mathrm{eq}}=\frac{3+(-9)}{2} \] \[ x_{\mathrm{eq}}=\frac{-6}{2}=-3\,\text{cm} \] Therefore, the equilibrium position does not coincide with the coordinate origin. \(\textbf{Amplitude calculation:}\) Amplitude is the maximum magnitude of displacement measured from equilibrium, not necessarily from \(x=0\). \[ A=\frac{x_{\max}-x_{\min}}{2} \] \[ A=\frac{3-(-9)}{2} \] \[ A=\frac{12}{2}=6\,\text{cm} \] The distance from equilibrium to either extreme is consequently \(6\,\text{cm}\). The full separation between the extreme positions is \(12\,\text{cm}\), which equals \(2A\). \(\textbf{Physical check:}\) \[ -3-6=-9\,\text{cm} \] \[ -3+6=+3\,\text{cm} \] Both calculated extremes agree with the supplied coordinates. \(\textbf{Final answer:}\) \[ \boxed{(x_{\mathrm{eq}},A)=(-3\,\text{cm},\,6\,\text{cm})} \]

Explanation: \(\textbf{Initial condition:}\) The particle begins at the positive extreme. \[ x_i=+A \] A traversal from one extreme to the opposite extreme represents half of a complete oscillation. \(\textbf{Tracking successive traversals:}\) After the first traversal, the particle reaches \(-A\). After the second traversal, it returns to \(+A\), completing one oscillation. After the third traversal, it reaches \(-A\) again. After the fourth traversal, it returns to \(+A\), completing two oscillations. After the fifth traversal, it reaches \(-A\). Thus, an odd number of extreme-to-extreme traversals leaves the particle at the extreme opposite to its starting position. \(\textbf{Number of oscillations:}\) \[ N=5\times\frac12 \] \[ N=\frac52=2.5 \] \(\textbf{Net displacement:}\) Displacement depends only on the initial and final coordinates. \[ \Delta x=x_f-x_i \] \[ \Delta x=(-A)-(+A)=-2A \] The negative sign indicates displacement towards the negative \(x\)-direction. \(\textbf{Distance check:}\) Each traversal covers a distance \(2A\). \[ s=5(2A)=10A \] The total distance \(10A\) differs from the net displacement \(-2A\) because the particle repeatedly reverses its direction. \(\textbf{Final answer:}\) \[ \boxed{\left(x_f,\Delta x,N\right)=\left(-A,-2A,\frac52\right)} \]

Explanation: Statement I is correct. In uniform circular motion, the moving point returns to the same position with the same velocity after each complete revolution. Therefore, its state repeats after a definite time interval, establishing periodic motion. However, the point travels continuously around the circumference rather than reversing repeatedly along a fixed diameter. Hence, periodic circular motion is not itself a linear to-and-fro oscillation. Statement II is also correct. The projection of a uniformly rotating point onto a fixed diameter moves repeatedly between the two ends of that diameter. The projected position passes through the centre in opposite directions during successive half-cycles. If the radius is \(R\) and the angular speed is \(\omega\), a suitable coordinate convention gives the projected displacement as \(x=R\cos(\omega t+\phi)\). This is sinusoidal motion with amplitude \(R\) and period \(2\pi/\omega\). The projected motion is, in fact, simple harmonic motion, even though the original point follows a circular path. The distinction arises because the actual path and the projected one-dimensional motion are different. Both statements are therefore correct.

Explanation: The pendulum continues moving alternately on either side of its equilibrium position. It also reverses its direction at successive extreme positions. Hence, its motion remains oscillatory while this to-and-fro movement continues. However, strict periodicity requires the complete state of motion to repeat after a fixed interval. Because air resistance removes mechanical energy, the maximum angular displacement gradually decreases. Consequently, successive extreme positions are not identical. Even if the intervals between successive swings are approximately equal, the actual motion does not repeat exactly. The decreasing amplitude therefore prevents strict periodicity. Ideal simple harmonic motion requires a constant-amplitude sinusoidal displacement under the appropriate restoring-acceleration condition. A real pendulum with decreasing amplitude does not satisfy that ideal description throughout its motion. Thus, oscillatory behaviour can persist even when exact periodicity is absent.

Explanation: \(\textbf{Given acceleration relation:}\) \[ a=-(9\,\text{s}^{-2})x \] The coefficient of displacement is constant, and the acceleration is directed opposite to the displacement. Therefore, the acceleration always points towards the equilibrium position. \(\textbf{SHM condition:}\) The defining acceleration equation of simple harmonic motion is \[ a=-\omega^2x \] The negative sign represents restoring acceleration. Comparing the given equation with the SHM condition, \[ \omega^2=9\,\text{s}^{-2} \] Taking the positive angular-frequency magnitude, \[ \omega=\sqrt{9}\,\text{rad s}^{-1} \] \[ \omega=3\,\text{rad s}^{-1} \] \(\textbf{Frequency distinction:}\) Angular frequency and ordinary frequency are not numerically identical. Their relation is \[ \omega=2\pi f \] Therefore, \[ f=\frac{\omega}{2\pi} \] \[ f=\frac{3}{2\pi}\,\text{Hz} \] The numerical coefficient \(9\) corresponds to \(\omega^2\), not to \(\omega\) or \(f\). The acceleration equation establishes SHM because it satisfies both proportionality to displacement and the restoring-direction requirement. \(\textbf{Final answer:}\) \[ \boxed{\omega=3\,\text{rad s}^{-1}} \]

Explanation: The particle occupies the same position \(P\) at both specified instants. However, its velocity directions are opposite. Therefore, the two observations represent different states of motion. For periodic motion, the complete state must repeat after each integral number of time periods. In particular, the position and velocity must both return to their earlier values. The state at \(t_2\) cannot be identical to the state at \(t_1\) because the velocities have opposite signs. Hence, their time separation cannot be an integral multiple of the fundamental period. It is also incorrect to conclude that the interval must always equal \(T/2\). For a sinusoidal oscillator, the separation between two passages through a general noncentral position in opposite directions depends on the position and the order of the passages. The special half-period separation applies to successive crossings of the equilibrium position in opposite directions. This distinction shows why repeated position alone is insufficient for identifying a complete oscillation.

Explanation: \(\textbf{Initial state:}\) The particle begins at the positive extreme. For SHM, the journey from one extreme to the opposite extreme takes half a period. \[ t_{\text{first negative extreme}}=\frac{T}{2} \] After reaching the negative extreme, the particle must complete one full oscillation before reaching that same extreme again. \(\textbf{Successive negative-extreme arrivals:}\) The first arrival occurs at \[ t_1=\frac{T}{2} \] The second arrival occurs at \[ t_2=\frac{3T}{2} \] The third arrival occurs at \[ t_3=\frac{5T}{2} \] \(\textbf{Using the observation:}\) \[ \frac{5T}{2}=5\,\text{s} \] \[ T=\frac{2(5)}{5}\,\text{s} \] \[ T=2\,\text{s} \] \(\textbf{Frequency calculation:}\) \[ f=\frac{1}{T} \] \[ f=\frac{1}{2\,\text{s}}=0.5\,\text{Hz} \] \(\textbf{Verification:}\) With \(T=2\,\text{s}\), successive arrivals at the negative extreme occur at \(1\,\text{s}\), \(3\,\text{s}\) and \(5\,\text{s}\). This agrees with the third-arrival observation. Counting the third arrival as \(3T/2\) would incorrectly ignore the initial half-period offset. \(\textbf{Final answer:}\) \[ \boxed{T=2\,\text{s},\qquad f=0.5\,\text{Hz}} \]

Explanation: \(\textbf{Observed crossings:}\) At \(t=0\), the particle crosses equilibrium in the positive direction. At \(t=0.6\,\text{s}\), it crosses equilibrium in the negative direction. At \(t=1.2\,\text{s}\), it again crosses equilibrium in the positive direction. \(\textbf{Identifying a complete oscillation:}\) Successive equilibrium crossings in SHM occur at intervals of \(T/2\). \[ \frac{T}{2}=0.6\,\text{s} \] Therefore, \[ T=2(0.6)\,\text{s} \] \[ T=1.2\,\text{s} \] The crossings at \(t=0\) and \(t=1.2\,\text{s}\) have the same position and velocity direction, so they represent repeated motion states. \(\textbf{Counting complete cycles:}\) \[ N=\frac{\text{Total elapsed time}}{\text{Time period}} \] \[ N=\frac{3.6\,\text{s}}{1.2\,\text{s}} \] \[ N=3 \] \(\textbf{Physical check:}\) At \(t=3.6\,\text{s}\), the particle again crosses equilibrium in the positive direction. Thus, its final motion state is the same as its initial state. The elapsed interval contains exactly three complete oscillations, not six; six would count successive half-cycles as complete cycles. \(\textbf{Final answer:}\) \[ \boxed{T=1.2\,\text{s},\qquad N=3} \]

Explanation: \(\textbf{Initial-condition equation:}\) Since the particle starts from its positive extreme at \(t=0\), its displacement can be expressed as \[ x=A\cos(\omega t) \] The angular frequency and period are related by \[ \omega=\frac{2\pi}{T} \] \(\textbf{Time to reach }x=A/2\textbf{:}\) Substituting the first specified position, \[ \frac{A}{2}=A\cos(\omega t_1) \] \[ \cos(\omega t_1)=\frac12 \] During the first journey towards equilibrium, \[ \omega t_1=\frac{\pi}{3} \] Therefore, \[ t_1=\frac{\pi}{3\omega} \] Using \(\omega=2\pi/T\), \[ t_1=\frac{\pi}{3}\times\frac{T}{2\pi} \] \[ t_1=\frac{T}{6} \] \(\textbf{Time to reach equilibrium:}\) Starting from the positive extreme, the first arrival at equilibrium occurs after one-quarter period. \[ t_{\mathrm{eq}}=\frac{T}{4} \] The second required interval is measured from \(x=A/2\) to \(x=0\). \[ t_2=t_{\mathrm{eq}}-t_1 \] \[ t_2=\frac{T}{4}-\frac{T}{6} \] \[ t_2=\frac{T}{12} \] \(\textbf{Ratio and physical interpretation:}\) \[ \frac{t_1}{t_2}=\frac{T/6}{T/12}=2 \] Although both path segments have length \(A/2\), the times are unequal. The particle has lower speed near the extreme and gains speed while approaching equilibrium. It therefore takes longer to cover the first segment. \(\textbf{Final answer:}\) \[ \boxed{t_1:t_2=2:1} \]

Explanation: \(\textbf{Initial motion condition:}\) The particle begins at equilibrium and moves towards the positive extreme. Therefore, its displacement can be represented by \[ x=A\sin(\omega t) \] \(\textbf{Finding the period:}\) The first journey from equilibrium to the positive extreme takes one-quarter period. \[ \frac{T}{4}=0.30\,\text{s} \] \[ T=4(0.30)=1.20\,\text{s} \] \(\textbf{Angular frequency:}\) \[ \omega=\frac{2\pi}{T} \] \[ \omega=\frac{2\pi}{1.20} \] \[ \omega=\frac{5\pi}{3}\,\text{rad s}^{-1} \] \(\textbf{Displacement at }t=0.45\,\text{s}\textbf{:}\) The phase accumulated since the initial equilibrium crossing is \[ \theta=\omega t \] \[ \theta=\frac{5\pi}{3}(0.45) \] \[ \theta=\frac{3\pi}{4} \] Substituting into the displacement equation, \[ x=A\sin\left(\frac{3\pi}{4}\right) \] \[ x=\frac{A}{\sqrt{2}} \] The displacement is positive, so the particle is still on the positive side of equilibrium. \(\textbf{Direction of motion:}\) The velocity is obtained from the displacement equation. \[ v=A\omega\cos(\omega t) \] At the specified instant, \[ v=A\omega\cos\left(\frac{3\pi}{4}\right) \] \[ v=-\frac{A\omega}{\sqrt{2}} \] The negative velocity indicates motion towards decreasing \(x\). Since the particle is at positive displacement, this means it is moving back towards equilibrium. This is consistent with the fact that it reached the positive extreme at \(0.30\,\text{s}\) and subsequently reversed direction. \(\textbf{Final answer:}\) \[ \boxed{x=+\frac{A}{\sqrt{2}},\qquad v=-\frac{A\omega}{\sqrt{2}}} \]

Explanation: \(\textbf{Interpretation of observations:}\) The two particles are observed for different durations. Therefore, their oscillation counts cannot be compared directly without accounting for elapsed time. Frequency is the number of complete oscillations per second. \[ f=\frac{N}{t} \] \(\textbf{Frequency of particle }P\textbf{:}\) \[ f_P=\frac{48}{24} \] \[ f_P=2\,\text{Hz} \] \(\textbf{Frequency of particle }Q\textbf{:}\) \[ f_Q=\frac{54}{18} \] \[ f_Q=3\,\text{Hz} \] \(\textbf{Comparing time periods:}\) Time period is the reciprocal of ordinary frequency. \[ T=\frac{1}{f} \] Therefore, \[ T_P=\frac12\,\text{s} \] \[ T_Q=\frac13\,\text{s} \] The ratio becomes \[ \frac{T_P}{T_Q}=\frac{1/2}{1/3} \] \[ \frac{T_P}{T_Q}=\frac32 \] \(\textbf{Comparing angular frequencies:}\) The angular frequency is \[ \omega=2\pi f \] Hence, \[ \omega_P=2\pi(2)=4\pi\,\text{rad s}^{-1} \] \[ \omega_Q=2\pi(3)=6\pi\,\text{rad s}^{-1} \] Therefore, \[ \frac{\omega_P}{\omega_Q}=\frac{4\pi}{6\pi}=\frac23 \] \(\textbf{Physical consistency:}\) Particle \(Q\) completes more oscillations per second, so it must have a smaller time period and a larger angular frequency. The obtained ratios satisfy both conditions. \(\textbf{Final answer:}\) \[ \boxed{T_P:T_Q=3:2,\qquad \omega_P:\omega_Q=2:3} \]

Explanation: \(\textbf{Phase advancement:}\) For SHM, the phase increases uniformly with time. The phase change over an interval is \[ \Delta\theta=\omega\Delta t \] \(\textbf{Finding angular frequency:}\) Given, \[ \Delta\theta=\frac{3\pi}{5}\,\text{rad} \] \[ \Delta t=0.15\,\text{s} \] Therefore, \[ \omega=\frac{\Delta\theta}{\Delta t} \] \[ \omega=\frac{3\pi/5}{0.15} \] Writing \(0.15=3/20\), \[ \omega=\frac{3\pi}{5}\times\frac{20}{3} \] \[ \omega=4\pi\,\text{rad s}^{-1} \] \(\textbf{Selecting the displacement equation:}\) The particle starts from the positive extreme, so \[ x=A\cos(\omega t) \] At \(t=0.375\,\text{s}\), \[ \theta=(4\pi)(0.375) \] \[ \theta=\frac{3\pi}{2} \] \(\textbf{Finding displacement:}\) \[ x=A\cos\left(\frac{3\pi}{2}\right) \] \[ x=0 \] The particle is therefore at equilibrium. \(\textbf{Determining velocity direction:}\) For cosine-form SHM, \[ v=-A\omega\sin(\omega t) \] Substituting the phase, \[ v=-A\omega\sin\left(\frac{3\pi}{2}\right) \] \[ v=+A\omega \] The velocity is positive and has its maximum magnitude. \(\textbf{Physical verification:}\) The period is \[ T=\frac{2\pi}{4\pi}=0.5\,\text{s} \] The specified time is \(3T/4\). Starting from the positive extreme, the particle reaches the negative extreme after \(T/2\) and crosses equilibrium moving positively after \(3T/4\). This confirms the displacement and velocity signs. \(\textbf{Final answer:}\) \[ \boxed{x=0,\qquad v=+A\omega} \]

Explanation: \(\textbf{Frequency relationship:}\) The given frequencies satisfy \[ f_Q=\frac32 f_P \] Since period and frequency are reciprocal, \[ T_P=\frac{1}{f_P},\qquad T_Q=\frac{1}{f_Q} \] Consequently, \[ \frac{T_Q}{T_P}=\frac{f_P}{f_Q} \] \[ \frac{T_Q}{T_P}=\frac23 \] Therefore, \[ T_Q=\frac23 T_P \] \(\textbf{Elapsed time for }P\textbf{:}\) Particle \(P\) first returns to its positive extreme after one complete period. \[ t=T_P \] \(\textbf{Oscillations completed by }Q\textbf{:}\) During this interval, the number of oscillations performed by \(Q\) is \[ N_Q=\frac{t}{T_Q} \] \[ N_Q=\frac{T_P}{(2/3)T_P} \] \[ N_Q=\frac32 \] Thus, \(Q\) performs one and a half oscillations. \(\textbf{Tracking equilibrium crossings:}\) Initially, \(Q\) is at \(+A\). After one-quarter of its period, it crosses equilibrium towards the negative side. After three-quarters of its period, it crosses equilibrium towards the positive side. After five-quarters of its period, it crosses equilibrium towards the negative side again. These are three distinct equilibrium crossings. \(\textbf{Final position of }Q\textbf{:}\) After one complete oscillation, \(Q\) returns to \(+A\). The additional half oscillation takes it from \(+A\) to \(-A\). Therefore, at \(t=3T_Q/2=T_P\), particle \(Q\) is at its negative extreme. The two particles are then at opposite extremes despite having started together. \(\textbf{Final answer:}\) \[ \boxed{N_{\mathrm{crossings}}=3,\qquad x_Q=-A} \]

Explanation: \(\textbf{Interpreting graph slopes:}\) The slope of a displacement-time graph represents instantaneous velocity. At \(t=0.20\,\text{s}\), the slope is negative, so the particle crosses equilibrium towards negative displacement. At \(t=0.50\,\text{s}\), the slope is positive, so it crosses equilibrium towards positive displacement. The crossings are consecutive because no other zero crossing occurs between them. \(\textbf{Finding the period:}\) In SHM, consecutive equilibrium crossings in opposite directions are separated by half a period. \[ \frac{T}{2}=0.50-0.20 \] \[ \frac{T}{2}=0.30\,\text{s} \] Therefore, \[ T=0.60\,\text{s} \] \(\textbf{Frequency calculation:}\) \[ f=\frac{1}{T} \] \[ f=\frac{1}{0.60} \] \[ f=\frac53\,\text{Hz} \] \(\textbf{Finding the state at }t=0.65\,\text{s}\textbf{:}\) The particle crosses equilibrium in the positive direction at \(t=0.50\,\text{s}\). From an equilibrium crossing towards positive displacement, the positive extreme is reached after \(T/4\). \[ \frac{T}{4}=\frac{0.60}{4} \] \[ \frac{T}{4}=0.15\,\text{s} \] Hence, \[ t_{\mathrm{positive\ extreme}}=0.50+0.15 \] \[ t_{\mathrm{positive\ extreme}}=0.65\,\text{s} \] \(\textbf{Physical interpretation:}\) At the positive extreme, the displacement is maximum and positive. The instantaneous velocity is zero because the particle reverses direction there. The negative-to-positive equilibrium crossing therefore determines which extreme is reached after the following quarter-period. \(\textbf{Final answer:}\) \[ \boxed{f=\frac53\,\text{Hz},\qquad x(0.65\,\text{s})=+A} \]

Explanation: \(\textbf{Identifying the component periods:}\) The given displacement is the sum of two sinusoidal functions. \[ x(t)=2\sin(2\pi t)+\sin(4\pi t) \] The first component has angular frequency \[ \omega_1=2\pi\,\text{rad s}^{-1} \] Its period is \[ T_1=\frac{2\pi}{\omega_1} \] \[ T_1=1\,\text{s} \] The second component has angular frequency \[ \omega_2=4\pi\,\text{rad s}^{-1} \] Therefore, \[ T_2=\frac{2\pi}{4\pi}=0.5\,\text{s} \] \(\textbf{Finding the common period:}\) After \(1\,\text{s}\), both components complete integral numbers of cycles. The first component completes one cycle, while the second completes two. Therefore, \[ x(t+1)=x(t) \] Thus, \(1\,\text{s}\) is a period of the complete function. \(\textbf{Checking the smaller candidate:}\) After \(0.5\,\text{s}\), the first sine term changes sign, while the second returns to its original value. \[ x(t+0.5)=-2\sin(2\pi t)+\sin(4\pi t) \] This is not generally equal to \(x(t)\). Since the two nonzero sinusoidal components have frequencies \(1\,\text{Hz}\) and \(2\,\text{Hz}\), their sum has fundamental period \(1\,\text{s}\). \(\textbf{Testing the SHM condition:}\) For SHM, the acceleration must satisfy \[ a=-\Omega^2x \] with one constant angular frequency \(\Omega\). Differentiating the given displacement twice with respect to time, \[ a(t)=-2(2\pi)^2\sin(2\pi t)-(4\pi)^2\sin(4\pi t) \] The two displacement components have different acceleration coefficients, namely \((2\pi)^2\) and \((4\pi)^2\). Therefore, the total acceleration cannot be expressed as \(-\Omega^2x\) using one constant value of \(\Omega\) for all times. The displacement is periodic, but it does not represent a single simple harmonic motion. \(\textbf{Final answer:}\) The fundamental period is \(1\,\text{s}\), and the motion is periodic but not simple harmonic.

Explanation: \(\textbf{Motion between the reflecting walls:}\) The particle repeatedly travels between two fixed extreme positions and reverses direction at each wall. The separation between the walls is \[ L=18-(-18) \] \[ L=36\,\text{cm} \] The speed between collisions is \[ v=18\,\text{cm s}^{-1} \] \(\textbf{Time for one complete cycle:}\) The particle starts at the left wall immediately after reflection. To regain the same position and direction, it must travel to the right wall and return to the left wall. Therefore, the distance travelled in one complete cycle is \[ s=2L \] \[ s=2(36)=72\,\text{cm} \] The time period is \[ T=\frac{s}{v} \] \[ T=\frac{72}{18}=4\,\text{s} \] \(\textbf{Locating the particle at }t=5\,\text{s}\textbf{:}\) After \(4\,\text{s}\), the particle returns to the left wall and again begins moving right. The remaining time is \[ \Delta t=5-4=1\,\text{s} \] During this interval, it travels \[ \Delta x=v\Delta t \] \[ \Delta x=18(1)=18\,\text{cm} \] Starting from \(x=-18\,\text{cm}\), \[ x=-18+18=0\,\text{cm} \] Thus, it is at the centre and moving towards the right. \(\textbf{Testing simple harmonicity:}\) The motion is oscillatory because it involves repeated to-and-fro movement between two extremes. It is also periodic because its complete state repeats every \(4\,\text{s}\). However, its acceleration is zero between collisions, while its velocity reverses instantaneously at the walls. This does not satisfy the continuous restoring-acceleration relation \(a=-\omega^2x\) of SHM. Therefore, the motion is periodic and oscillatory but not simple harmonic. \(\textbf{Final answer:}\) The period is \(4\,\text{s}\), and the particle is at the centre moving right at \(t=5\,\text{s}\). The motion is not SHM.

Explanation: \(\textbf{Period of the actual oscillation:}\) The given displacement equation is \[ x(t)=A\sin(5\pi t) \] Comparing it with \(x=A\sin(\omega t)\), \[ \omega=5\pi\,\text{rad s}^{-1} \] The particle's time period is \[ T_x=\frac{2\pi}{\omega} \] \[ T_x=\frac{2\pi}{5\pi}=0.4\,\text{s} \] \(\textbf{Period of the measured quantity:}\) The sensor measures the square of displacement. \[ S(t)=x^2(t) \] \[ S(t)=A^2\sin^2(5\pi t) \] Using the trigonometric identity \[ \sin^2\theta=\frac{1-\cos(2\theta)}{2} \] we obtain \[ S(t)=\frac{A^2}{2}[1-\cos(10\pi t)] \] The oscillatory part of the sensor reading has angular frequency \(10\pi\,\text{rad s}^{-1}\). Therefore, \[ T_S=\frac{2\pi}{10\pi} \] \[ T_S=0.2\,\text{s} \] \(\textbf{Identifying successive sensor maxima:}\) The sensor reading is maximum whenever \[ x^2=A^2 \] This occurs at both \(x=+A\) and \(x=-A\). The first sensor maximum after \(t=0\) occurs when \[ 5\pi t=\frac{\pi}{2} \] \[ t=0.1\,\text{s} \] At this instant, \[ x=A \] The next sensor maximum occurs at \(t=0.3\,\text{s}\). \[ x=A\sin(5\pi\times0.3) \] \[ x=A\sin\left(\frac{3\pi}{2}\right)=-A \] Thus, consecutive sensor maxima represent opposite extreme positions. The square removes the information about the sign of displacement, causing the recorded signal to repeat twice during one actual oscillation. \(\textbf{Final answer:}\) \[ \boxed{T_S=0.2\,\text{s},\qquad x_1=+A,\quad x_2=-A} \]

Explanation: \(\textbf{Interpreting the two passages:}\) Both observations occur at the same positive displacement. However, the velocity is negative during the first passage and positive during the second. The particle must travel towards the negative extreme, reverse direction, and then return to \(x=A/2\). The two instants therefore represent different phases of the same oscillation. \(\textbf{Selecting a phase representation:}\) Write the displacement as \[ x=A\cos\theta \] where the phase \(\theta\) increases uniformly with time. At \(x=A/2\), \[ \cos\theta=\frac12 \] The two relevant phases within one cycle are \[ \theta_1=\frac{\pi}{3} \] and \[ \theta_2=\frac{5\pi}{3} \] At \(\theta_1=\pi/3\), the velocity is negative because \(v=-A\omega\sin\theta\). At \(\theta_2=5\pi/3\), the velocity is positive. These phase values therefore match the specified directions. \(\textbf{Phase change between the passages:}\) \[ \Delta\theta=\theta_2-\theta_1 \] \[ \Delta\theta=\frac{5\pi}{3}-\frac{\pi}{3} \] \[ \Delta\theta=\frac{4\pi}{3} \] \(\textbf{Determining the time period:}\) The elapsed time is \[ \Delta t=0.50-0.10=0.40\,\text{s} \] Using \(\Delta\theta=\omega\Delta t\), \[ \frac{4\pi}{3}=\omega(0.40) \] \[ \omega=\frac{10\pi}{3}\,\text{rad s}^{-1} \] Therefore, \[ T=\frac{2\pi}{\omega} \] \[ T=\frac{2\pi}{10\pi/3} \] \[ T=0.60\,\text{s} \] \(\textbf{Finding the next positive extreme:}\) At \(t=0.50\,\text{s}\), the phase is \(5\pi/3\). The next positive extreme occurs at phase \(2\pi\). The remaining phase change is \[ \Delta\theta' =2\pi-\frac{5\pi}{3}=\frac{\pi}{3} \] Consequently, \[ \Delta t'=\frac{\Delta\theta'}{\omega} \] \[ \Delta t'=\frac{\pi/3}{10\pi/3} \] \[ \Delta t'=0.10\,\text{s} \] The positive extreme is reached at \[ t=0.50+0.10=0.60\,\text{s} \] \(\textbf{Physical check:}\) The interval between the two specified passages equals \(2T/3\), not \(T/2\), because the position is not the equilibrium position. \(\textbf{Final answer:}\) \[ \boxed{T=0.60\,\text{s},\qquad t_{\mathrm{next\ positive\ extreme}}=0.60\,\text{s}} \]

Explanation: \(\textbf{General displacement relation:}\) The particle executes SHM, so its displacement may be expressed as \[ x=A\cos(\omega t+\phi) \] The initial displacement determines the cosine of the phase constant. \[ x(0)=A\cos\phi \] Given \(x(0)=A/2\), \[ A\cos\phi=\frac{A}{2} \] \[ \cos\phi=\frac12 \] Within one cycle, this condition permits \[ \phi=\frac{\pi}{3} \] or \[ \phi=\frac{5\pi}{3} \] \(\textbf{Using the initial velocity direction:}\) Differentiating the displacement equation, \[ v=-A\omega\sin(\omega t+\phi) \] At \(t=0\), \[ v(0)=-A\omega\sin\phi \] The initial velocity is negative. Therefore, \(\sin\phi\) must be positive. This selects \[ \phi=\frac{\pi}{3} \] The correct displacement equation is consequently \[ x=A\cos\left(\omega t+\frac{\pi}{3}\right) \] \(\textbf{Time to the negative extreme:}\) The negative extreme occurs when \[ x=-A \] Therefore, \[ \cos\left(\omega t+\frac{\pi}{3}\right)=-1 \] The first suitable phase is \(\pi\). \[ \omega t+\frac{\pi}{3}=\pi \] \[ \omega t=\frac{2\pi}{3} \] \[ t=\frac{2\pi}{3\omega} \] Since \(T=2\pi/\omega\), \[ t=\frac{T}{3} \] \(\textbf{Consistency check:}\) The particle initially lies on the positive side and is moving towards the negative side. It must first pass equilibrium and then reach the negative extreme. The positive phase branch correctly describes this motion, whereas the negative phase branch initially gives positive velocity. \(\textbf{Final answer:}\) \[ \boxed{x=A\cos\left(\omega t+\frac{\pi}{3}\right),\qquad t_{\mathrm{first\ negative\ extreme}}=\frac{T}{3}} \]