SOME SYSTEMS EXECUTING SIMPLE HARMONIC MOTION -

OSCILLATING OF BODY ATTACHED WITH SPRING

CASE I-HORIZONTAL

Consider a block of mass m attached with horizontal spring of spring constant k is oscillate on the frictionless surface as shown in the figure.

In the equilibrium position, there are no restoring force acts. If we displace the block through a small distance x to the right, then according to Hook’s law the spring exerts a restoring force on the block to the left and given by.

$\fn_cm \large F=-kx\;where\;k\;is\;force/spring\;constant$

$\fn_cm \large \Rightarrow ma=-kx$

$\fn_cm \large \Rightarrow a=-\frac{k}{m}x$

we know that, if the system oscillates, then it follow $\fn_cm \large a=-\omega^2 x$

i.e the system oscillates about the mean position

On comparing, we get $\fn_cm \large \omega^2=\frac{k}{m}$

$\fn_cm \large \Rightarrow \omega=\sqrt{\frac{k}{m}}$

The time period of oscillation is given by

$\fn_cm \large T=\frac{2\pi}{\omega}$

$\fn_cm \large \left [ T=2\pi\sqrt{\frac{m}{k}}\; \right ]$

Its frequency is given by

$\fn_cm \large \left [ f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{m}{k}}\; \right ]$

CASE II- VERTICAL

Consider a body of mass m is suspended vertically with a spring of spring constant k. Due to load, the spring stretches and come to rest in an equilibrium position (mean position). In this case, gravity shifts the equilibrium position.

$\fn_cm \large k \Delta x=mg----------(1)$

If the mass is displaced downward by an amount x, then the net restoring force exerted on the body in the upward direction is

$\fn_cm \large F=-[k(x+\Delta x)-mg]$

$\fn_cm \large \Rightarrow F=-[kx+k\Delta x-mg]$

$\fn_cm \large \Rightarrow F=-kx$

$\fn_cm \large \Rightarrow ma=-kx$

$\fn_cm \large \Rightarrow a=-\frac{k}{m}x$

we know that, if the system oscillates, then it follow $\fn_cm \large a=-\omega^2 x$

i.e the system oscillates about the mean position

On comparing, we get $\fn_cm \large \omega^2=\frac{k}{m}$

$\fn_cm \large \Rightarrow \omega=\sqrt{\frac{k}{m}}$

The time period of oscillation is given by

$\fn_cm \large T=\frac{2\pi}{\omega}$

$\fn_cm \large \left [ T=2\pi\sqrt{\frac{m}{k}}\; \right ]$

Its frequency is given by

$\fn_cm \large \left [ f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{m}{k}}\; \right ]$

from (1) $\fn_cm \large k=\frac{mg}{\Delta x}$

$\fn_cm \large \therefore T=2\pi\sqrt{\frac{m}{\frac{mg}{\Delta x}}}$

$\fn_cm \large \left [ T=2\pi\sqrt{\frac{\Delta x}{g}} \right ]$

Q.1> A 5 kg collar is attached to a spring of spring constant 500 N m–1. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate
(a) the period of oscillation,
(b) the maximum speed and
(c) maximum acceleration of the collar.

Q.2> A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Q.3> A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Q.4> In the Previous question, let us take the position of mass when the spring is unstreched 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(a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

SIMPLE PENDULUM

Consider a simple pendulum of mass of bob is m and length of the string is l. Let any time string makes an angle θ with the vertical.

From fig, $\fn_cm \large mg\sin\theta$ serves as a restoring force that tends to bring the bob to the equilibrium position.

The torque due to this force is

$\fn_cm \large \vec{\tau}=\vec{r}\times\vec{F}$

$\fn_cm \large \Rightarrow \tau=[lmg \sin\theta]\sin 90^0$

$\fn_cm \large \Rightarrow \tau=-mg\sin\theta l$

As we know torque is $\fn_cm \large \tau=I\alpha$ where $\fn_cm \large I$ is the moment of inertia of the bob and $\fn_cm \large \alpha$ is its angular acceleration

$\fn_cm \large I\alpha =-mg\sin\theta l$

$\fn_cm \large \Rightarrow ml^2\alpha=-mg\sin\theta l$

$\fn_cm \large \alpha=-\frac{g}{l}\sin\theta$

for small angle $\fn_cm \large \sin\theta\approx \theta$

$\fn_cm \large \therefore \alpha=-\frac{g}{l}\theta$

we know that for angular S.H.M

$\fn_cm \large \alpha=-\omega^2 \theta$

$\fn_cm \large i.e\;\;\omega^2=\frac{g}{l}\;\;\;\;\;\;\;\therefore \omega=\sqrt{\frac{g}{l}}$

The time period of oscillation is

$\fn_cm \large T=\frac{2\pi}{\omega}\;\;\;\;\;\;\left [ T=2\pi\sqrt{\frac{l}{g}} \right ]$

Its frequency is

$\fn_cm \large \left [ f=\frac{1}{2\pi}\sqrt{\frac{g}{l}} \right ]$

NOTE

The correct equation of the time period of a simple pendulum is

$\fn_cm \large T=2\pi\sqrt{\frac{l}{g_{eff}}}$

In a free-falling lift, $\fn_cm \large g_{eff}=0\;\;and\;\;T=\infty$ i.e in this case pendulum will not oscillate.
If the lift is at rest or moving down/up with constant velocity, then $\fn_cm \large T=2\pi\sqrt{\frac{l}{g}}$
If the lift is moving upward with constant acceleration a, then $\fn_cm \large T=2\pi\sqrt{\frac{l}{g+a}}$
If the lift is moving downward with constant acceleration a, then $\fn_cm \large T=2\pi\sqrt{\frac{l}{g-a}}$

The restoring force in a simple pendulum is $\fn_cm \large F=mg\sin\theta$ . In arriving at the formula $\fn_cm \large T=2\pi\sqrt{\frac{l}{g}}$ , we take $\fn_cm \large \sin\theta\approx \theta$ i.e restoring force $\fn_cm \large F=mg\theta$ . For a large value of $\fn_cm \large \theta,\sin\theta<\theta$ . Therefore restoring force decreases from $\fn_cm \large mg\theta\;to\; mg\sin\theta$ . As a result Time period increases and the frequency decreases.

Q. 5> What is the length of a simple pendulum, which ticks seconds?

Q. 6> The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 $\fn_cm \large m/s^2$ )

Q. 7> Answer the following questions :

(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle: $\fn_cm \large T=2\pi \sqrt{\frac{m}{k}}$ . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than $\fn_cm \large 2\pi \sqrt{\frac{l}{g}}$ . Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity ?

Q. 8>A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

OSCILLATION OF A LIQUID IN A U-TUBE

Consider a liquid filled in a vertical U-tube of uniform cross-sectional area up to a height h as shown in the figure. In the equilibrium position, the liquid level in both arms is the same.

Suppose we push liquid through a distance Y, the liquid of the other arm rises up the same distance. As a result in one arm, there is an additional liquid column of length 2Y.

Here the weight of the liquid column of height 2Y serves as the restoring force.

Let $\fn_cm \large A\rightarrow$ Area of the cross-section of the tube

$\fn_cm \large \rho\rightarrow$ Density of liquid

The restoring force is $\fn_cm \large F=weight\;of\;liquid$

$\fn_cm \large \Rightarrow F=-\rho(A2y)g\;\;\;\;\;\;\;\;\;\;(\rho=\frac{m}{v})$

$\fn_cm \large \Rightarrow F=-\rho A 2g y$

$\fn_cm \large \because a=\frac{F}{m}=-\frac{\rho A2gy}{\rho A2h}$

$\fn_cm \large \therefore a=-\left ( \frac{g}{h} \right )y$

we know that for S.H.M

$\fn_cm \large a=\omega^2 y$

$\fn_cm \large \therefore \omega=\sqrt{\frac{g}{h}}$

Therefore the time period of oscillation is

$\fn_cm \large T=\frac{2\pi}{\omega}$

$\fn_cm \large \left [ T=2\pi \sqrt{\frac{h}{g}} \; \right ]$

and its frequency is $\fn_cm \large \left [ f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{g}{h}} \right ]$

OSCILLATION OF A FLOATING CYLINDER

Consider a wooden cylinder of length $\fn_cm \large l$ and cross-sectional area $\fn_cm \large A$ floating in a liquid of density $\fn_cm \large \rho_l$ . Let the density of cylinder is $\fn_cm \large \rho_b$ .

In equilibrium position, let the cylinder is floating with a depth $\fn_cm \large h$ submerged. In this situation, the upward buoyancy force by liquid is balanced by the weight of the cylinder.

If the cylinder is pushed down a small distance $\fn_cm \large y$ and then release, it will move up and down with SHM. Here the extra buoyancy force exerted by the liquid in an upward direction serves as a restoring force.

i.e $\fn_cm \large F=weight\;of\;extra\;displaced\;liquid$

$\fn_cm \large \Rightarrow F=-\rho_lAyg$

$\fn_cm \large \because a=\frac{F}{m}$

$\fn_cm \large \therefore a=-\frac{\rho_l Ayg}{\rho_b Al}$

$\fn_cm \large \Rightarrow a=-\left ( \frac{\rho_l g}{\rho_b l} \right )y$

We know that for S.H.M

$\fn_cm \large a=-\omega^2 y$

i.e $\fn_cm \large \omega^2=\frac{\rho_l g}{\rho_b l}$

$\fn_cm \large \therefore \omega=\sqrt{\frac{\rho_l g}{\rho_b l}}$

Hence time period of oscillation is $\fn_cm \large T=\frac{2\pi}{\omega}$

$\fn_cm \large \left [ T=2\pi \sqrt{\frac{\rho_b l}{\rho_l g}} \right ]$

OSCILLATION OF A BALL IN THE NECK OF AN AIR CHAMBER

From fig, if the ball is depressed by distance $\fn_cm \large y$ , decrease in volume, $\fn_cm \large \Delta V=A \times y$

If $\fn_cm \large \Delta p$ is the increase in pressure applied on the ball, then, bulk modulus of elasticity of air is

$\fn_cm \large K=- \frac{\Delta p}{\Delta V/V}=-\frac{\Delta p}{Ay/V}$

$\fn_cm \large \therefore \Delta p=-\frac{KA}{V}y$

Restoring force, $\fn_cm \large F=\Delta p \times A=-\frac{KAy}{V}\times A=-\frac{KA^2}{V}y$

Since F is directly proportional to $\fn_cm \large y$ and acts in opposite direction, the ball execute S.H.M

$\fn_cm \large \because a=\frac{F}{m}$

$\fn_cm \large \therefore a=-\frac{KA^2}{mV}y$

we know that for S.H.M

$\fn_cm \large a=-\omega ^2 y$

$\fn_cm \large i.e\;\;\; \omega ^2=\frac{KA^2}{mV}$

$\fn_cm \large \therefore \omega=\sqrt{\frac{KA^2}{mV}}$

Hence time period of oscillation is $\fn_cm \large T=\frac{2\pi}{\omega}$

$\fn_cm \large \left [ T=2\pi \sqrt{\frac{mV}{KA^2}} \right ]$

If P-V variations are isothermal, then $\fn_cm \large K=\Delta p$

$\fn_cm \large \therefore \left [ T=2\pi\sqrt{\frac{mV}{\Delta p A^2}} \right ]$

If P-V variations are adiabatic, then $\fn_cm \large K=\gamma \Delta p$

$\fn_cm \large i.e \left [ T=2\pi\sqrt{\frac{mV}{\gamma \Delta p A^2}} \right ]$

TWO BODY OSCILLATOR (HORIZONTAL)

consider two mass $\fn_cm \large m_1$ and $\fn_cm \large m_2$ attached with a spring of spring constant K. Let l is the natural length of the spring. The system is oscillating on a horizontal frictionless surface.

Here no external force acts on the system to oscillate. i.e C.O.M is at rest. Let C.O.M is situated at point O.

Then we know that from the concept of COM

$\fn_cm \large l_1=\frac{m_2l}{m_1 +m_2}\;\;\;\;\;and\;\;\;\;\;l_2=\frac{m_1 l}{m_1 +m_2}$

Now, $\fn_cm \large k_1=\frac{k l}{l_1}\;\;\;\;\;and\;\;\;\;\;k_2=\frac{kl}{l_2}$

$\fn_cm \large \therefore k_1=\frac{kl(m_1 +m_2)}{m_2 l}\;\;\;\;\;and\;\;\;\;\;k_2=\frac{kl(m_1 +m_2)}{m_1 l}$

$\fn_cm \large \therefore k_1=\frac{k(m_1 +m_2)}{m_2}\;\;\;\;\;and\;\;\;\;\;k_2=\frac{k(m_1 +m_2)}{m_1}$

The time period of $\fn_cm \large m_1\;and \;m_2$ with respect to COM is the same which is the same time period of the system.

i.e $\fn_cm \large T=2\pi \sqrt{\frac{m_1}{k_1}}\;\;\;\;\;or\;\;\;\;\;T=2\pi \sqrt{\frac{m_2}{k_2}}$

$\fn_cm \large \left [ T=2\pi \sqrt{\frac{m_1 m_2}{k(m_1 +m_2)}} \right ]$

if $\fn_cm \large m_1 =m_2=m$ then

$\fn_cm \large T=2\pi \sqrt{\frac{m^2}{k 2 m}}$

$\fn_cm \large \left [ T=2\pi \sqrt{\frac{m}{2k}} \right ]$

Q. 9> Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.30 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.30(b) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases ?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ?

Q. 10> The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

TORSIONAL PENDULUM

In a torsion pendulum, the pendulum tends to perform rotational motion about the vertical axis of the wire, which results in the twisting of the wire rather than swinging like a normal pendulum. The force which twists the pendulum keeps reversing its direction to maintain equilibrium. Thus this pendulum also follows the necessary condition for SHM that the acceleration/force is opposite in direction to the displacement. The pendulum performing such oscillations is known as the Torsional pendulum. In this pendulum the restoring force is torque.

If the angular displacement of the body at an instant is θ, the resultant torque acting on the body in angular simple harmonic motion should be $\fn_cm \large \tau =-k \theta$

If the moment of inertia is $I$ , the angular acceleration is

$\fn_cm \large \alpha =\frac{\tau }{I}$

$\fn_cm \large \Rightarrow \alpha =-\frac{k}{I}\theta$

We know that for angular simple harmonic motion

$\fn_cm \large \alpha=-\omega^2 \theta$

$\fn_cm \large \therefore \omega^2=\frac{k}{I}\;\;\;\Rightarrow \;\;\; \omega=\sqrt{\frac{k}{I}}$

The time period of the oscillation is given by

$\fn_cm \large T=\frac{\2\pi}{\omega}$

$\fn_cm \large \left [ T=2\pi \sqrt{\frac{I}{k}} \right ]$

Q.11> A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = –α θ , where J is the restoring couple and θ the angle of twist).

IMPORTANT LINKS OF THIS CHAPTER

INTRODUCTION OF OSCILLATIONS	PERIODIC AND OSCILLATORY MOTIONS
SIMPLE HARMONIC MOTION	SIMPLE HARMONIC MOTION AND UNIFORM CIRCULAR MOTION
VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION	FORCE LAW FOR SIMPLE HARMONIC MOTION
ENERGY IN SIMPLE HARMONIC MOTION	SOME SYSTEMS EXECUTING SIMPLE HARMONIC MOTION
DAMPED SIMPLE HARMONIC MOTION	FORCED OSCILLATIONS AND RESONANCE