Q.In Exercise 13.9, let us take the position of mass when the spring is unstretched as $x = 0$, and the direction from left to right as the positive direction of $x$-axis. Give $x$ as a function of time $t$ for the oscillating mass if at the moment we start the stopwatch ($t = 0$), the mass is

The key idea is that two pendulums released together from the same extreme will be in phase again when each has completed a whole number of oscillations — and those whole numbers must be such that the elapsed time is the same for both. That time is simply the least common multiple (LCM) of their time periods.

Let’s work it out.

1. Find the time periods. For a simple pendulum, $T = 2\pi \sqrt{\frac{L}{g}}$. Given $g = \pi^2$ m/s²,

2. What “in phase” means here. Both start from the same extreme position (say, maximum displacement to the right). They are in phase when they again reach that same extreme together. That happens after a time $t$ such that

where $n_1$ and $n_2$ are positive integers (the number of oscillations each has completed). We want the smallest such $t$.

3. Find the smallest common multiple of the periods. Write the periods as fractions:

The condition $n_1 \cdot 2 = n_2 \cdot \frac{12}{5}$ gives

For $n_1$ to be an integer, $n_2$ must be a multiple of 5. The smallest is $n_2 = 5$, giving $n_1 = 6$.

Then

That gives 12 s — but wait, that’s not the smallest possible time. Let’s check more carefully.

4. Re-examine the meaning.

   If they only need to be at the same point in their cycle (same displacement and same direction of motion), then the condition is $t = n_1 T_1 = n_2 T_2$ as above. That gave 12 s.

   But the problem says “in phase when released from the same extreme position” — so they start together at an extreme. The next time they are both at that same extreme together is indeed after the LCM of the periods.

   However, there is a nuance: they could also be in phase at the other extreme (opposite displacement but same direction of motion relative to equilibrium). That would also count as being in phase? Usually, “in phase” means same displacement and same velocity direction. Starting from the right extreme, being at the left extreme together is also in phase (both at extreme, both about to move toward the mean). That would happen after half the LCM? Let’s check.

   At $t = T_1/2 = 1$ s, pendulum 1 is at the left extreme. At $t = T_2/2 = 1.2$ s, pendulum 2 is at the left extreme. Not the same time.

   At $t = 3$ s, pendulum 1 has done 1.5 oscillations — back at left extreme. Pendulum 2 at $t=3$ s has done $3/2.4 = 1.25$ oscillations — not at an extreme. So no. …

Concept & Intuition

In simple harmonic motion, the restoring force is proportional to displacement and opposite in direction: $F = -k y$. The given equation $F + 0.04\pi^2 y = 0$ is exactly of that form, so we can identify $k = 0.04\pi^2\ \mathrm{N/m}$. The angular frequency $\omega$ for a mass $m$ is $\omega = \sqrt{k/m}$. Once we have $\omega$, the maximum velocity during SHM is simply $v_{\max} = \omega A$, where $A$ is the amplitude. The key is to be careful with units — mass is given in grams, so convert to kilograms.

Step-by-step solution

1. Convert mass to SI units

   The mass is $90\ \mathrm{g} = 0.09\ \mathrm{kg}$. Always work in kg, m, s.

2. Identify the force constant

   The equation is $F + 0.04\pi^2 y = 0$, i.e. $F = -0.04\pi^2 y$.

   Comparing with $F = -k y$, we get

3. Find angular frequency $\omega$ For SHM, $\omega = \sqrt{\frac{k}{m}}$. Substitute:

4. Use amplitude to get maximum velocity …

The key insight is that the equation $F + 0.04\pi^2 y = 0$ is simply the standard SHM force law $F = -k y$ written in a slightly disguised form. In simple harmonic motion, the restoring force is always proportional to the displacement and directed opposite to it. Here, rearranging gives $F = -0.04\pi^2 y$, so the force constant $k$ is $0.04\pi^2$ N/m.

Once we have $k$ and the mass $m$, the angular frequency $\omega$ follows from $\omega = \sqrt{k/m}$. And for any SHM, the maximum velocity is $v_{\max} = A\omega$, where $A$ is the amplitude. The problem gives both $m$ and $A$ explicitly, so it becomes a straightforward substitution.

1. Identify the force constant.

   The given equation is $F + 0.04\pi^2 y = 0$. Rewrite it as $F = -0.04\pi^2 y$.

   Comparing with $F = -k y$, we get $k = 0.04\pi^2$ N/m.

2. Convert mass to kilograms.

   Mass is 90 g = $0.09$ kg. Always use SI units in such problems.

3. Find angular frequency $\omega$.

   For SHM, $\omega = \sqrt{\frac{k}{m}}$.

4. Apply the maximum velocity formula. …

The motion is $y = 8\cos(\pi t)\ \text{cm}$. "Displacement in the time interval" means the change in position $\Delta y = y(2.5) - y(1.5)$.

At $t = 1.5\ \text{s}$:

At $t = 2.5\ \text{s}$:

Therefore …

Energy budget. This oscillator already stores 5 J of potential energy at the mean position, so that part is a constant background. The energy that genuinely oscillates is

Spring constant. At an extreme the whole oscillatory energy is elastic:

$E_{\text{osc}}=\tfrac12 kA^2\ \Rightarrow\ 4=\tfrac12 k(0.01)^2=\tfrac12 k\times10^{-4},$ …

The problem gives you a time period of 16 s, so the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{16} = \frac{\pi}{8}$ rad/s. That part is fixed. The challenge is that you don't know the phase constant — the particle passes through the origin at t = 2 s, but that doesn't mean it's at the extreme position or at maximum velocity; it just means displacement is zero at that instant.

The natural approach: write the displacement as $x = A\sin(\omega t + \phi)$ or $x = A\cos(\omega t + \phi')$. Either works, but sine is convenient because at the origin, $\sin(\theta) = 0$ gives a clean condition. Let's use $x = A\sin(\omega t + \phi)$.

1. Use the condition at t = 2 s At t = 2 s, $x = 0$, so:

Since $A \neq 0$, we have $\sin\left(\frac{\pi}{4} + \phi\right) = 0$.

This means $\frac{\pi}{4} + \phi = n\pi$, where $n$ is an integer. The simplest choice is $n = 0$, giving $\phi = -\frac{\pi}{4}$. (Other integer values just shift the time origin; they won't change the amplitude we find.)

2. Write the velocity equation Velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. With $\omega = \frac{\pi}{8}$ and $\phi = -\frac{\pi}{4}$, we have:

3. Use the condition at t = 4 s At t = 4 s, $v = 4$ m/s. Substitute:

Simplify the angle: $\frac{\pi}{8} \cdot 4 = \frac{\pi}{2}$, so the argument is $\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

Thus: …

The motion of a particle executing Simple Harmonic Motion (SHM) is characterized by its oscillation about a mean position. The key to solving problems involving SHM is understanding how its kinematic quantities – position, velocity, and acceleration – are related to each other and to the fundamental parameters of the motion: amplitude ($A$) and angular frequency ($\omega$).

The defining characteristic of SHM is that the restoring force, and thus the acceleration, is directly proportional to the displacement from the mean position and directed towards it.

Here are the essential formulas for SHM:

Our strategy will be to use the given information to form a system of equations involving $\omega$ and $A$, and then solve for $A$.

1. Use the acceleration information to find $\omega^2$: We are given that the acceleration of the particle is $3\ \mathrm{cm/s^2}$ when its distance from the mean position is $1\ \mathrm{cm}$. Using the magnitude of the acceleration formula, $|a| = \omega^2 |x|$: $3\ \mathrm{cm/s^2} = \omega^2 (1\ \mathrm{cm})$ This directly gives us the value of $\omega^2$:

2. Use the velocity information to form an equation involving $\omega$ and $A$: …

The problem gives you the displacement equation $y = 6 \sin(100t - \pi/4)$ cm. The key is to recognize that in SHM, kinetic energy is maximum when the particle passes through the mean position, where all the energy is kinetic. That maximum kinetic energy is exactly the total mechanical energy of the oscillator, because at the extreme positions all energy is potential, and at the mean all energy is kinetic.

So you don’t need to differentiate or find velocity explicitly — just use the formula for total energy in SHM: $E = \frac{1}{2} m \omega^2 A^2$, where $A$ is amplitude and $\omega$ is angular frequency.

1. Extract amplitude and angular frequency.

   The equation is $y = 6 \sin(100t - \pi/4)$ cm. Compare with the standard form $y = A \sin(\omega t + \phi)$.

   Amplitude $A = 6$ cm. Angular frequency $\omega = 100$ rad/s.

   The phase constant $-\pi/4$ doesn’t affect energy.

2. Convert units consistently.

   Mass is given in kg, so work in SI units. Convert amplitude to metres:

   $A = 6\,\text{cm} = 0.06\,\text{m}$.

3. Apply the energy formula.

   Maximum kinetic energy $K_{\text{max}} = \frac{1}{2} m \omega^2 A^2$.

   Substitute $m = 1\,\text{kg}$, $\omega = 100\,\text{rad/s}$, $A = 0.06\,\text{m}$: …

The key idea is that in simple harmonic motion, the total mechanical energy is conserved and splits between kinetic energy (KE) and potential energy (PE) in a way that depends on displacement. The problem gives a specific ratio — KE is eight times PE — and asks for the corresponding displacement from the mean position.

Why does this approach work? Because both KE and PE have simple expressions in terms of displacement $y$ and amplitude $A$:

• PE at displacement $y$ is $\frac{1}{2} m \omega^2 y^2$
• KE at displacement $y$ is $\frac{1}{2} m \omega^2 (A^2 - y^2)$
• Total energy $E = \frac{1}{2} m \omega^2 A^2$

The ratio condition lets you solve directly for $y$ in terms of $A$, without needing $m$, $\omega$, or the phase constant. The given equation $y = \sqrt{3\pi} \sin\left( \frac{100}{\pi} t + \frac{\pi}{4} \right)$ tells us the amplitude $A = \sqrt{3\pi}$.

1. Write the energy condition. We are told KE = 8 × PE. Using the standard SHM energy expressions:

Cancel $\frac{1}{2} m \omega^2$ (non-zero) from both sides:

2. Solve for $y^2$. Bring terms together:

So $y = \pm \frac{A}{3}$. The magnitude of displacement is $A/3$.

3. Read the amplitude from the given equation. The displacement is $y = \sqrt{3\pi} \sin\left( \frac{100}{\pi} t + \frac{\pi}{4} \right)$. …

When a wave travels, different points in space experience the wave's oscillation at different times. The "phase" of a wave at a particular point and instant describes its position in its cycle of oscillation. The "phase difference" between two points tells us how much one point is ahead or behind the other in its oscillatory cycle.

For a wave propagating in a medium, points separated by one full wavelength ($\lambda$) are exactly in phase, meaning their oscillations are perfectly synchronized. Points separated by half a wavelength are exactly out of phase. This relationship is linear: the phase difference is directly proportional to the path difference between the two points.

The key idea is that a path difference of $\lambda$ corresponds to a phase difference of $2\pi$ radians (or $360^\circ$). Therefore, for any path difference $\Delta x$, the phase difference $\Delta \phi$ can be found using the ratio:

This gives us the fundamental relationship:

Here's how to apply this to the given problem:

1. Identify the given quantities:

   We are given the frequency of the sound wave, $f = 200 \text{ Hz}$.

   The speed of sound in air is $v = 340 \text{ m/s}$.

   The separation between the two points (path difference) is $\Delta x = 85 \text{ cm}$.

2. Convert units for consistency:

   The speed is in meters per second, so we should convert the path difference from centimeters to meters:

   $\Delta x = 85 \text{ cm} = 0.85 \text{ m}$.

3. Calculate the wavelength ($\lambda$):

   The relationship between wave speed ($v$), frequency ($f$), and wavelength ($\lambda$) is given by:

We can rearrange this to find the wavelength:

Substituting the given values: …

The key here is that the three waves are all at the same angular frequency $\omega$, so they can be added using the phasor (or vector) method. Each wave is represented as a vector whose length is the amplitude and whose angle is the phase relative to a reference. The resultant amplitude is the magnitude of the vector sum.

Let’s work through it step by step.

1. Write all waves in cosine form with a consistent phase reference.

   We take $x = A \cos(\omega t)$ as the reference (phase $0$).

   • $x_1 = A \cos \omega t$ → amplitude $A$, phase $0$.
   • $x_2 = 2A \sin \omega t = 2A \cos\left(\omega t - \frac{\pi}{2}\right)$ → amplitude $2A$, phase $-\frac{\pi}{2}$.
   • $x_3 = \sqrt{2}A \cos\left(\omega t + \frac{\pi}{4}\right)$ → amplitude $\sqrt{2}A$, phase $+\frac{\pi}{4}$.

2. Represent each as a phasor (vector) in the complex plane.

   The real axis corresponds to phase $0$, the imaginary axis to phase $\pm \frac{\pi}{2}$.

   • Phasor of $x_1$: $A$ along the positive real axis → $(A, 0)$.
   • Phasor of $x_2$: $2A$ at $-\frac{\pi}{2}$ → along the negative imaginary axis → $(0, -2A)$.
   • Phasor of $x_3$: $\sqrt{2}A$ at $+\frac{\pi}{4}$ → components: Real part: $\sqrt{2}A \cos\frac{\pi}{4} = \sqrt{2}A \cdot \frac{1}{\sqrt{2}} = A$ Imaginary part: $\sqrt{2}A \sin\frac{\pi}{4} = \sqrt{2}A \cdot \frac{1}{\sqrt{2}} = A$ So phasor $= (A, A)$.

3. Add the phasors component-wise.

   • Total real part: $A + 0 + A = 2A$ …