In the previous section, we saw specific heats of gases. In this chapter, we will see oscillations.
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In some earlier chapters, we have seen rectilinear motion, projectile motion and uniform circular motion.
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In uniform circular motion, a particle takes the same time to complete each revolution. So it is a periodic motion.
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In this chapter, we will see another type of periodic motion, called oscillation.
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In oscillation, a particle moves to and fro about a mean position.
♦ Motion of the pendulum is an example
♦ Motion of the piston inside the cylinder of an engine is another example.
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To learn about periodic motion and oscillatory motion, we must first see some fundamental concepts like period, frequency, displacement etc., They are explained below:
Periodic and Oscillatory motions
Some basics can be written in 5 steps:
1. Consider an insect climbing up a wall. It starts climbing from the ground level. It travels with a uniform speed u1. In a time duration of t1, it reaches a height h. This is shown by the red line in fig.14.1(a) below:
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| Fig.14.1 |
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After t1, the insect is not able to climb upwards. Neither is it able to hold on. So it climbs down to the ground with a uniform speed of u2. The time taken for reaching the ground is t2. This is shown by the green line in fig.14.1(a) above.
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Note that, t1 is larger than t2. This is because, time taken to reach the ground is smaller.
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The sum (t1 + t2) is T. This T is the time required to start from the ground and reach back to the ground. If the insect keep repeating the "climbing and falling" process, we will get several repetitions in the graph. The ground is then considered as the mean position.
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The red line is familiar to us. We have plotted many distance-time (s-t) graphs in rectilinear motion. If the motion is uniform, we get the same red line. It is the plot of the equation s = ut. In our present case, u1 is the slope of the red line. u1 is constant.
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The green line can also be given a similar explanation. It has a negative slope because, velocity is negative for downward motion. So slope of the green line is −u2.
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In fig.14.1, instead of s, we plot x(t) along the y-axis. This is because, in periodic motion, the displacement from the mean position is denoted by x. It is a function of time. So we write x(t).
2. Consider the game of bouncing the ball between palm and ground.
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The game starts when the ball is thrown to the ground from a height h. It is thrown with an initial speed of u1. Since the ball is under the influence of gravity, speed is not uniform.
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The displacement at any instant 't' is given by:
$\small{x(t)\,=\,u_1 t + \frac{1}{2} a t^2}$. This is the equation of a parabola.
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In a time duration of t1, it reaches the ground. This is shown by the red parabola in fig.14.1(b) above.
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After t1, the ball bounces upwards with a speed of u2. Since the ball is under the influence of gravity, speed is not uniform. The displacement at any instant 't' is given by:
$\small{x(t)\,=\,u_2 t - \frac{1}{2} a t^2}$. This is the equation of a parabola.
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In a time duration of t2, it reaches the palm. This is shown by the green parabola in fig.14.1(b) above.
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The sum (t1 + t2) is T. This T is the time required to start from the palm and reach back to the palm. As the game continues, we will get several repetitions in the graph. The ground is then considered as the mean position.
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Note that, the red and green parabolas are graphs. They are not the paths followed by the ball. The path of the ball is always vertical. While playing this game, the ball cannot be thrown in a parabolic path. If the ball is thrown in a parabolic path, it will not bounce back to the palm. It will bounce away from the player.
3. A motion that repeats itself at regular intervals of time is called a periodic motion.
4. Consider the red sphere in fig.14.2 below. It is at the bottom of the yellow bowl.
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| Fig.14.2 |
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In this position, the sphere is at equilibrium. If the sphere is left alone, it will remain there forever.
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But if the sphere is given a displacement, a force will come into play, which tries to bring the sphere back to the equilibrium position. As a result, the sphere will perform oscillation or vibration in the bowl.
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Oscillating bodies eventually come to rest. In our present case, the sphere will come to rest at the bottom of the bowl.
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This is due to the friction between the sphere and the bowl. The surrounding air also causes the sphere to come to rest. This type of oscillation is called damped oscillation.
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However, an oscillation can be forced to remain oscillating, by providing an external agency. This type of oscillation is called forced oscillation. We will see the details of both types in later sections of this chapter.
5. Every oscillatory motion is a periodic motion.
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But every periodic motion need not be an oscillatory motion.
♦ Circular motion is periodic but not oscillatory.
Period and frequency
This can be explained in 7 steps:
1. Consider the earlier fig.14.1(a). We turn on the stop-watch at the instant when the insect begins to climb up. When the reading in the stop-watch is T, the insect is back on the ground, and the next cycle begins.
2. In fig.14.1(b), we turn on the stop-watch at the instant when the ball is thrown. When the reading in the stop-watch is T, the ball is back at the palm, and the next cycle begins.
3. So we can write:
The smallest duration of time after which the next cycle begins, is called the period of the periodic motion. It is denoted by the symbol T.
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The S.I unit of period is second.
4. For periodic motions which are too fast, period will be very small. In such cases, instead of s, we can use other units.
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For example, the period of vibration of a quartz crystal is expressed in microseconds.
♦ One microsecond is $\small{10^{-6}}$ seconds.
♦ Microsecond is abbreviated as: $\small{{\rm{\mu s}}}$
5. For periodic motions which are too slow, period will be very large. In such cases also, instead of s, we can use other units.
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For example, the orbital period of some planets are expressed in earth days.
♦ For mercury, it is 88 earth days.
♦ One earth day is 24 hours.
6. Suppose that, period of a periodic motion is 30 s. Then in one minute, two cycles will occur.
♦ That is., in 60 s, two cycles will occur
♦ That is., in 1 s, 2/60 cycles will occur.
♦ That is., in 1 s, 1/30 cycles will occur.
♦ That is., in 1 s, 1/30 of one cycle will occur.
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The number of cycles in one second is called frequency. It is represented by the symbol $\small{\nu}$. In our present case, $\small{\nu}$ = 1/30.
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Note that, 1/30 is the reciprocal of 30. And 30 is the period in seconds.
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So we can write:
Frequency is the reciprocal of period. The period must be in seconds.
♦ That is., $\small{\nu~=~\frac{1}{T}}$
♦ where T is n seconds.
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Since we are taking the reciprocal of "period in seconds", the unit of frequency is $\small{{\rm{s^{-1}}}}$.
This $\small{{\rm{s^{-1}}}}$ is given a special name. It is called hertz (abbreviated as Hz), in honor of the scientist Heinrich Rudolph Hertz, who discovered radio waves.
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We can write:
1 hertz = 1 Hz = 1 cycle per second = 1 $\small{{\rm{s^{-1}}}}$.
7. Note that, $\small{\nu}$ need not be an integer. It can be a decimal or a fraction.
Let us see a solved example:
Solved example 14.1
On an average, a human heart is found to beat 75 times in a minute. Calculate it's frequency and period.
Solution:
1. First, let us calculate the frequency:
Number of occurrences in one minute = 75
⇒ Number of occurrences in 60 s = 75
⇒ Number of occurrences in 1 s = 75/60
⇒ Frequency $\small{\nu~=~\frac{75}{60}~=~\frac{5}{4}~=~1.25~{\rm{s^{-1}}}}$
2. Now we can calculate the period:
We know that, frequency is the reciprocal of period. It follows that, period is the reciprocal of frequency. So we can write:
Period $\small{T~=~\frac{1}{\nu}~=~\frac{4}{5}~=~0.8 {\rm{s}}}$
In the next
section, we will see displacement.
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