Q.Four displacement-time ($x$-$t$) plots for the linear motion of a particle are described below. In each plot $x$ is on the vertical axis and time $t$ (in seconds) on the horizontal axis. State which plots represent periodic motion and give the period in each periodic case.

Concept and Intuition

Any transcendental function (sin, cos, log, exponential) can only take a pure (dimensionless) number as its argument, and the function's output is likewise dimensionless. This is the standard trick for finding the dimension of a quantity that sits inside such a function.

Step-by-Step Solution

1. $\cos(\alpha t)$ is dimensionless $\Rightarrow \alpha t$ is dimensionless $\Rightarrow [\alpha]=\dfrac{1}{[t]}=T^{-1}=M^0L^0T^{-1}$.
2. Since $\cos(\alpha t)$ contributes no dimension, $[\beta]=\left[\dfrac{F}{v^2}\right]$.
3. $[F]=MLT^{-2}$, $[v^2]=(LT^{-1})^2=L^2T^{-2}$.
4. $[\beta]=\dfrac{MLT^{-2}}{L^2T^{-2}}=ML^{-1}T^0$.

Common Mistakes

• Assigning $\alpha$ dimension $T^0$ (forgetting it must cancel $t$'s dimension inside the cosine).
• Mis-simplifying $L/L^2=L^{-1}$ as $L^{1}$ or dropping the $M$.

ANSWER: C

Concept and Intuition

Sound travels as a wave train; the number of wave crests present in a stretch of space of length $d$ at any instant is simply $d/\lambda$, where $\lambda = v/f$ is the wavelength. Since both source and observer are stationary, there is no Doppler shift — the wavelength in the medium is just $v/f$ throughout.

Step-by-Step Solution

1. Wavelength: $\lambda = \dfrac{v}{f} = \dfrac{340}{680} = 0.5\,\mathrm{m}$.
2. Number of waves between source and observer: $n = \dfrac{d}{\lambda} = \dfrac{150}{0.5} = 300$.

Common Mistakes

• Trying to invoke a Doppler formula even though both source and observer are stationary (no relative motion, no shift needed).
• Inverting the ratio ($\lambda/d$ instead of $d/\lambda$).

ANSWER: A

Concept and Intuition

A one-dimensional travelling wave is written as $y(x,t)=A\sin(kx-\omega t)$, where $k=\dfrac{2\pi}{\lambda}$ is the wave number (spatial repetition rate) and $\omega = 2\pi f$ is the angular frequency (temporal repetition rate). These two quantities describe two independent periodicities of the wave — one in space, one in time — and only the time-periodicity coefficient tells you the frequency.

Step-by-Step Solution

1. Write the given wave: $y(x,t) = 0.5\sin(70.1x - 10\pi t)$.
2. Match with the standard form $y = A\sin(kx - \omega t)$: here $k = 70.1\ \mathrm{rad/m}$ and $\omega = 10\pi\ \mathrm{rad/s}$.
3. Frequency is obtained purely from $\omega$: $f = \dfrac{\omega}{2\pi} = \dfrac{10\pi}{2\pi} = 5\ \mathrm{Hz}$.

Common Mistakes

• Trying to use the $70.1x$ coefficient (the wave number) to compute frequency — it only gives wavelength via $\lambda = 2\pi/k$, not frequency.
• Forgetting the factor of $2\pi$ and reporting $\omega$ itself (i.e., $10\pi \approx 31.4$) as the frequency.

ANSWER: D

Concept and Intuition

A function's period is the smallest $T$ for which $f(t+T)=f(t)$ for all $t$. Squaring a sinusoid effectively doubles its angular frequency (halves the period), because $\sin^2\theta$ is always non-negative and repeats twice within one full cycle of $\sin\theta$. This is captured algebraically by the power-reduction identity.

Step-by-Step Solution

1. Use the identity $\sin^2\omega t=\dfrac{1-\cos(2\omega t)}{2}$.
2. This is a constant ($1/2$) plus a cosine term of angular frequency $\omega' = 2\omega$.
3. The period of $\cos(\omega' t)$ is $T=\dfrac{2\pi}{\omega'}=\dfrac{2\pi}{2\omega}=\dfrac{\pi}{\omega}$.
4. Adding the constant $1/2$ does not change the period, so the period of $\sin^2\omega t$ is $\dfrac{\pi}{\omega}$ seconds.

Common Mistakes

• Assuming the period of $\sin^2\omega t$ is the same as that of $\sin\omega t$, i.e. $2\pi/\omega$ — this ignores the frequency-doubling effect of squaring.
• Misapplying the double-angle formula and getting a $\sqrt{\ }$ in the answer (there is no square root in a period expression here — those options are distractors).

ANSWER: B

Concept and Intuition

Two masses connected through a spring network oscillate about their common centre of mass exactly like a single mass $\mu$ (the reduced mass) attached to a spring of the network's effective stiffness. So the first job is to reduce the spring network to one number, $k_{eff}$; the second is to apply the standard two-body oscillator frequency $\omega=\sqrt{k_{eff}/\mu}$.

Step-by-Step Solution

1. The two springs of constant $K$ between the left block and the rigid middle junction act in parallel (same displacement): combined constant $=2K$.
2. This combination is in series (through the massless junction) with the single spring $K$ connecting to the right block, so $\dfrac{1}{k_{eff}}=\dfrac{1}{2K}+\dfrac{1}{K}=\dfrac{3}{2K}\Rightarrow k_{eff}=\dfrac{2K}{3}$.
3. With $K=9\pi^2$: $k_{eff}=\dfrac{2\times9\pi^2}{3}=6\pi^2\ \mathrm{Nm^{-1}}$.
4. Reduced mass for two equal 3 kg blocks: $\mu=\dfrac{m_1m_2}{m_1+m_2}=\dfrac{3\times3}{6}=1.5$ kg.
5. Angular frequency: $\omega=\sqrt{k_{eff}/\mu}=\sqrt{6\pi^2/1.5}=\sqrt{4\pi^2}=2\pi\ \mathrm{rad\,s^{-1}}$.
6. Time period: $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2\pi}=1$ s.

Common Mistakes

• Adding the springs as if all three were in one simple series or parallel chain, without recognising the parallel pair first.
• Using the total mass ($6$ kg) instead of the reduced mass in a two-body oscillator problem.

ANSWER: A

Concept and Intuition

The current $i(t)=3\sin(\omega t)$ rises from 0 to a peak of 3 A at $\omega t=\pi/2$, then falls back through the rms value and down to 0 at $\omega t=\pi$. We need the point on this falling part of the cycle where $i$ equals the rms value $3/\sqrt2$, then measure the time from there to the next zero.

Step-by-Step Solution

1. RMS value of $i=3\sin(\omega t)$ is $i_{rms}=\dfrac{3}{\sqrt2}$.
2. Set $3\sin(\omega t)=\dfrac{3}{\sqrt2}\Rightarrow\sin(\omega t)=\dfrac{1}{\sqrt2}$. This is satisfied at $\omega t=\pi/4$ (rising) and $\omega t = \pi-\pi/4=3\pi/4$ (falling, i.e. after the peak).
3. Since the current is dropping from rms to zero, we take the falling-branch instant: $\omega t_1=3\pi/4$.
4. The next zero of the sine after the peak occurs at $\omega t_2=\pi$.
5. Time to drop from rms to zero: $\Delta t = \dfrac{\omega t_2-\omega t_1}{\omega}=\dfrac{\pi-3\pi/4}{\omega}=\dfrac{\pi/4}{\omega}=\dfrac{\pi}{4\omega}$.

Common Mistakes

• Using the rising-branch instant ($\omega t=\pi/4$) instead of the falling one, which would (incorrectly) describe current rising to rms, not dropping from it.

ANSWER: B

Concept and Intuition

Frequency of oscillation of any point on a wave equals the number of complete cycles it undergoes per unit time, independent of the wave's speed or wavelength (those relate to how far the disturbance travels, not how fast a single point oscillates).

Step-by-Step Solution

1. Convert time: one-fourth of a minute $= \dfrac{60}{4} = 15$ s.
2. Frequency $= \dfrac{\text{number of cycles}}{\text{time}} = \dfrac{120}{15} = 8\ \mathrm{Hz}$.

Common Mistakes

• Forgetting to convert "a fourth of a minute" into seconds before dividing.
• Confusing frequency of oscillation of a point with the wave's propagation speed.

ANSWER: A

The key insight is that period depends only on mass and the effective stiffness of the spring combination. In both configurations, the mass is the same, so the ratio of periods is simply the square root of the inverse ratio of effective spring constants.

Why this works:

For a single spring of constant $k$, the period is $T = 2\pi\sqrt{m/k}$. When springs are combined, we replace $k$ with an effective spring constant $k_{\text{eff}}$ that describes how the whole system responds to a force. The period then becomes $T = 2\pi\sqrt{m/k_{\text{eff}}}$. So if we find $k_{\text{eff}}$ for each configuration, we can directly compare their periods.

1. Find the effective spring constant for configuration (i) — series. When two identical springs are connected end-to-end (series), each spring stretches under the same force. The total extension is the sum of each spring’s extension. For a force $F$, each spring extends by $x = F/k$, so total extension is $2F/k$. The effective constant satisfies $F = k_{\text{eff, series}} \cdot (2F/k)$, giving

So the period for (i) is

We are told $T_i = 2\,\text{s}$.

2. Find the effective spring constant for configuration (ii) — parallel. When two identical springs are side-by-side (parallel), they share the load. For a force $F$ applied to the mass, each spring feels half the force, so each extends by $x = (F/2)/k = F/(2k)$. But both springs have the same extension, so the total extension is still $F/(2k)$. The effective constant satisfies $F = k_{\text{eff, parallel}} \cdot (F/(2k))$, giving

Hence the period for (ii) is

3. Relate the two periods. Compare $T_{ii}$ to $T_i$:

So $T_{ii} = \frac{1}{2} T_i = \frac{1}{2} \times 2\,\text{s} = 1\,\text{s}$.