Oscillations — NCERT Solutions

For the mass-spring system in Exercise 13.9, the amplitude is the maximum displacement, so
$$A=2.0\text{ cm}.$$
The angular frequency is
$$\omega=20\ \text{rad s}^{-1}.$$
Hence the SHM can be written as $x=A\cos(\omega t+\phi)$ with different $\phi$ depending on the initial position.

(a) At the mean position at $t=0$

Then $x(0)=0$, so we may take
$$x= A\sin \omega t = 2.0\sin(20t)\ \text{cm}.$$

(b) At the maximum stretched position at $t=0$

Then $x(0)=+A$, so
$$x=2.0\cos(20t)\ \text{cm}.$$

(c) At the maximum compressed position at $t=0$

Then $x(0)=-A$, so
$$x=-2.0\cos(20t)\ \text{cm}.$$

These functions have the same frequency and same amplitude, but differ in the initial phase.

The reference-circle construction gives the SHM as the projection of uniform circular motion. From each circular motion in Fig. 13.20, the corresponding x-projection is obtained by taking the x-coordinate of the rotating particle. The textbook method is:

1. Identify the radius of the circle as the amplitude. 2. Compute the angular speed using $\omega=2\pi/T$. 3. Read the initial position from the given starting point. 4. Use the sense of rotation to decide whether the projection is given by sine or cosine, with the correct phase. If you provide Fig. 13.20, I can write the exact SHM equations for both cases.

The corresponding reference-circle diagrams can be obtained using the form
$$x=A\cos(\omega t+\phi) \quad \text{or} \quad x=A\sin(\omega t+\phi).$$
For each motion:

• the radius of the circle equals the amplitude,
• the angular speed is $\omega$,
• the initial position at $t=0$ is given by the phase.

(a) $x=-2\sin(3t+\pi/3)$

Rewrite as
$$x=2\cos\left(3t+\frac{\pi}{3}+\frac{\pi}{2}\right)$$
or equivalently find phase from the sine form.

• Amplitude/radius: $2$ cm
• Angular speed: $3\ \text{rad s}^{-1}$
• Initial position:

$$x(0)=-2\sin\left(\frac{\pi}{3}\right)=-\sqrt{3}\ \text{cm}$$

(b) $x=\cos(\pi/6-t)$

Rewrite:
$$x=\cos\left(t-\frac{\pi}{6}\right)$$

• Amplitude/radius: $1$ cm
• Angular speed: $1\ \text{rad s}^{-1}$
• Initial position:

$$x(0)=\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt3}{2}\ \text{cm}$$

(c) $x=3\sin(2\pi t+\pi/4)$

• Amplitude/radius: $3$ cm
• Angular speed: $2\pi\ \text{rad s}^{-1}$
• Initial position:

$$x(0)=3\sin\left(\frac{\pi}{4}\right)=\frac{3\sqrt2}{2}\ \text{cm}$$

(d) $x=2\cos\pi t$

• Amplitude/radius: $2$ cm
• Angular speed: $\pi\ \text{rad s}^{-1}$
• Initial position:

$$x(0)=2\cos 0=2\ \text{cm}$$

For the reference circle, draw a circle of the given radius, mark the initial point corresponding to these $x(0)$ values, and show anticlockwise rotation with the stated $\omega$.

This is Exercise 13.13, which has two parts.

(a) Maximum extension of the spring

Case (a): one end fixed

If a force $F$ is applied to a spring of constant $k$, extension is
$$F=kx \Rightarrow x=\frac{F}{k}.$$
So the maximum extension is
$$\boxed{\frac{F}{k}}.$$

Case (b): both ends free, mass at each end

Each end is stretched by the same force $F$, so each half of the spring extends by
$$\frac{F}{k}.$$
Hence the total extension is
$$\boxed{\frac{2F}{k}}.$$

(b) Period of oscillation

Case (a)

For one mass attached to a spring,
$$T=2\pi\sqrt{\frac{m}{k}}.$$

Case (b)

Each mass experiences a restoring force proportional to displacement with effective spring constant doubled, so
$$T=2\pi\sqrt{\frac{m}{2k}}.$$

So:

• maximum extension: $F/k$ in case (a), $2F/k$ in case (b)
• period: $2\pi\sqrt{m/k}$ in case (a), $2\pi\sqrt{m/(2k)}$ in case (b)

For SHM, maximum speed is
$$v_{\max}=\omega A.$$
Stroke = twice amplitude = 1.0 m, so
$$A=0.5\text{ m}.$$
Angular frequency is
$$\omega=200\ \text{rad/min}=\frac{200}{60}\ \text{rad/s}=\frac{10}{3}\ \text{rad/s}.$$
Thus
$$v_{\max}=\omega A=\frac{10}{3}\times 0.5=\frac{5}{3}\ \text{m/s}.$$
So the maximum speed is about 1.67 m/s.

The book’s exercise text gives angular frequency in rad/min, so if the unit is kept as min-based speed, it is
$$200\times 0.5=100\ \text{m/min}=\frac{5}{3}\ \text{m/s}.$$

For a simple pendulum,
$$T=2\pi\sqrt{\frac{L}{g}}.$$
For the same pendulum, $L$ is constant, so
$$T\propto \frac{1}{\sqrt{g}}.$$
Therefore,
$$\frac{T_m}{T_e}=\sqrt{\frac{g_e}{g_m}}.$$
Given:
$$T_e=3.5\text{ s},\quad g_e=9.8\text{ m s}^{-2},\quad g_m=1.7\text{ m s}^{-2}.$$
So
$$T_m=3.5\sqrt{\frac{9.8}{1.7}}.$$
$$\frac{9.8}{1.7}\approx 5.765, \quad \sqrt{5.765}\approx 2.40.$$
Thus
$$T_m\approx 3.5\times 2.40\approx 8.4\text{ s}.$$

For small oscillations of the pendulum in the accelerating car, the effective restoring acceleration is due to the resultant of gravity and the car’s centripetal acceleration. The mean equilibrium position is along the resultant field.

If the car moves on a circular track of radius $R$ with speed $v$, its centripetal acceleration is
$$a_c=\frac{v^2}{R}.$$
The effective gravitational acceleration is then
$$g' = \sqrt{g^2 + a_c^2} = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2}.$$
For a simple pendulum of length $l$, the period of small oscillations is
$$T=2\pi\sqrt{\frac{l}{g'}}.$$
Hence
$$\boxed{T=2\pi\sqrt{\frac{l}{\sqrt{g^2+(v^2/R)^2}}}}.$$

Let the cork have base area $A$, height $h$, and density $\rho$. It floats in liquid density $\rho_l$.

If it is depressed by a small distance $x$, the volume of liquid displaced increases by $Ax$. So the extra upthrust is
$$F = \rho_l gAx,$$
directed upward.

This is a restoring force, so
$$F=-\rho_l gA x.$$
The mass of the cork is
$$m=\rho Ah.$$
Hence the equation of motion is
$$m\ddot x=-\rho_l gA x.$$
Substitute $m=\rho Ah$:
$$\rho Ah\,\ddot x=-\rho_l gA x.$$
Cancelling $A$,
$$\ddot x=-\frac{\rho_l g}{\rho h}x.$$
This is the equation of SHM with
$$\omega^2=\frac{\rho_l g}{\rho h}.$$
Therefore the period is
$$T=2\pi\sqrt{\frac{\rho h}{\rho_l g}}.$$
So the cork executes simple harmonic motion with period
$$\boxed{T=2\pi\sqrt{\frac{h\rho}{\rho_l g}}}.$$

Let the mercury column in the U-tube be displaced by a small distance $x$. Then one side rises by $x$ and the other side falls by $x$, so the difference in levels is $2x$.

The pressure difference is therefore
$$\Delta P=\rho g(2x),$$
where $\rho$ is the density of mercury.

If $A$ is the cross-sectional area of the tube, the restoring force on the mercury column is
$$F=-\Delta P\,A=-2\rho gAx.$$
The mass of the moving mercury column of length $l$ is
$$m=\rho Al.$$
Thus
$$m\ddot x=-2\rho gAx.$$
Substitute $m=\rho Al$:
$$\rho Al\,\ddot x=-2\rho gAx.$$
Cancelling $\rho A$,
$$\ddot x=-\frac{2g}{l}x.$$
This is the equation of SHM, with
$$\omega^2=\frac{2g}{l}.$$
Hence the period is
$$T=2\pi\sqrt{\frac{l}{2g}}.$$
Therefore, when the suction pump is removed, the mercury column executes simple harmonic motion.