**Simple harmonic motion**

**(A) illustration, explanation and definition of simple harmonic motion (S.H.M) **

A periodic motion (or vibration or oscillation)is a type of motion that occurs very frequently in physics. In such a motion, the path of a moving body is repeated at successive equal intervals of time.e.g motion of a body in a circular path with uniform speed, the oscillation of a clock pendulum, the vibrations of the balance wheel of a watch etc. A very important type of periodic motion is the simple harmonic motion, which can be defined thus:

*If a small body or a particle vibrates or moves to and fro along a straight line under the influence of a force so that its acceleration towards a fixed point(or its equilibrium position) is proportional to its distance or displacement from that point, the body is said to perform simple harmonic motion.*

**Illustrations of simple harmonic motion**

***Mass on a spring**

The spring with mass executing simple harmonic motion

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***The simple pendulum**

This consists of a small lead weight, or Bob, suspended from a rigid support by means of a thread.

The simple pendulum

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The period of motion** T** of the simple pendulum of length

*l*can be equal to

*T = 2π√l/g *

Where *g* is the acceleration due to gravity. This expression is valid provided that the angular displacement ~~O~~ or the amplitude of motion *a* is very small.

Therefore *T=√l*

**Experimental determination of ***g*** using simple pendulum **

The value of *g *can be determined experimentally by finding the time (t) taken for a number (n) of oscillations for different values of *l*, and hence obtained a table of time period **T =t/n** and ** l**.

Usually, n is between 20 and 50

Since we can rearrange the equation

*T = 2π√l/g, as l =g/4π*^{2}*T*^{2}

When performing the above experiment, certain precautions should be taken to obtain an accurate result.

- The amplitude of oscillation must be very small.
- The support for the cord must be rigid.
- The string should be very light and inextensible.
- Care should be taken to count from when the Bob passes the centre of the oscillation. A vertical mark arranged behind the thread when it is in its equilibrium position is helpful.

***Loaded test-tube in a liquid**

It can be demonstrated by adding sand into a test-tube. Put just enough sand into the test-tube so as to make it stand vertically erect in a liquid, e.g water.you will notice that while u depress the tube into the water and then release it, the tube moves up and down in the water in a regular fashion.

Loaded test-tube

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**(B)Speed and acceleration of S.H.M**

Speed or angular velocity (*w*) is given by

*W = *~~O~~*/t*

* = s/r*/t

* = s/t . 1/r*

* = V 1/r*

* = V/r*

* i.e. w. = V/r*

* **V = rw*

Linear velocity equals the product of the radius and the angular velocity.

**Acceleration of S.H.M**

Acceleration *p* may be written as:

*a*_{p}* = w*^{2}*xsin*~~O~~* = w*^{2}*r where r=xsin*~~O~~

**Relationship between angular acceleration and linear acceleration**

*Angular acceleration is the rate of change of angular velocity (w) with time (t).*

The unit of angular acceleration is radians per second (rads^{-2}). Suppose the angular velocity (*w*) of a body changed from *w*_{0}* to w*_{t}* in t sec., Then angular acceleration is given as *~~ O~~* = w*_{t}* – w*_{0}* / t*

* But w = v/r or v = rw*

*Hence *~~O~~* = 1/r (v*_{t}* – v*_{0}*)/t*

* Therefore *~~O~~* = a/r*

*Where a is the linear acceleration*

* ** a = *~~O~~*r*

Linear acceleration equals the product of angular acceleration and the radius (or displacement) of the particle from its central position.

**(C) period, frequency and amplitude if a body executing S.H.M**

**Period*** T: **this is the time required to make one complete revolution about a point of reference.* Period is measured in seconds (s).

*Frequency f is the number of complete revolutions per unit of time.* Frequency is measured in Hertz (H_{z}) or per second (s^{-1}). 1Hz = 1 cycle per second.

Frequency is the reciprocal of the period.

* **F = 1/T*

Where f is frequency, t is period.

*Note: **amplitude, A is defined as the maximum displacement of the body from the equilibrium position.*

**(D) Energy of simple harmonic motion**

The force necessary to stretch the spring a distance, *x *is given by. *F = kx*

Where *k *is the force constant or stiffness of the spring. The work done in stretching the spring is given by work = average force × x = ½ kx

Thus the energy stored in the stretched spring is ½ kx^{2}. Maximum energy stored or maximum potential energy (P.E) in a stretched spring is given by P.E_{max} = ½ kx. Where x is the maximum displacement or amplitude of the motion. This maximum energy is conserved throughout the simple harmonic motion.

The kinetic energy (K.E) at any instant of the motion is ½ mv^{2}, where m is the mass of the body and v is its velocity at that instant. When the oscillating spring is momentarily at rest at the position of maximum displacement, where v = 0 and kinetic energy equals zero, the potential energy equals ½ kx^{2}.

Total energy = K.E + P.E = ½ kx^{2} Total Energy at any point x, = ½ mv^{2} + ½ kx^{2}.

**Forced vibration and resonance**

**Resonance** is a vibration set up in a body due to a transfer of energy from another body which is vibrating with a period (or frequency) the same or nearly the same as that of the first.

**Note:** Resonance is a phenomenon which occurs whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses or signals received from some other system or body which is vibrating with the same frequency.

**Past questions**

- Which of the following is a stringed instrument? (Wassce 1989) A. Flute B. Trumpet C. Piano D. Drum E. Saxophone

**Answer:** C

**Solution:** piano is a stringed instrument which when touch vibrates to make sounds depending on it frequency

2. An alternating voltage with a frequency of 50Hz has a period of (wassce 1995)

A. 0.02s

B. 0.05s

C. 0.20s

D. 0.50s

E. 50.00s

**Answer:** A

**Solution:** f = 1/t

50 = 1/t

50t = 1

T = 1/50 = 0.02s

3. An object of mass 0.40kg attached to the end of a string is whirled round in a horizontal circle of radius 2.0m with a constant speed of 8ms^{-1}. Calculate the angular velocity of the object. (Wassce 1997)

A. 0.8rads^{-1}

B. 2.0rads^{-1}

C. 4.0rads^{-1}

D. 8.0rads^{-1}

E. 16.0rads^{-1}

**Answer:** C

**Solution:** v = rw =

*W* = ? r= 2.0m v =8m/s

v/r = 8/2 = 4rad/s

4. The Bob of a simple pendulum takes 8.0s to complete 10 oscillations . Determine the frequency of oscillation of the Bob.(wassce 2018)

A. 0.80Hz

B. 1.25Hz

C. 7.50Hz

D. 8.00Hz

**Answer:** B

**Solution:** t = 8.0s ; n = 10

f = 1/T = 1/t/n = n/t = 10/8 = 5/4

= 1.25Hz (Hertz)

5. The relationship between the period T and the length l of a simple pendulum is T =2π(l/g)^{½}. From experimental data of T and I, one can obtain the following graphs.

(I) T vs l (ii) T vs l^{2} (III) T^{2} vs l (IV) T vs √I (v) log T vs log l. Which of the graphs are linear ? (Jamb 1981)

.A. All

B. Only (I) and (ii)

C. Only (III) and (IV)

D. (III), (IV) and (v)

E. None of them

**Answer:** D

6. A simple pendulum 0.6m long, has a period of 1.5s. What is the period of a similar pendulum 0.4m long in the same location? (Jamb 1984)

A. 1.5√2/3s

B. 1.5√3/2s

C. 2.25s

D. 1.00s

E. 2.00s

**Answer:** A

**Solution:** T = 2π√l/g;. T_{1} /T_{2} = √l_{1 }/l_{2}

1 – 5/ T_{2} = √0.6/0.4; T_{2} =1.5√2/3s

7. The length of a displaced pendulum bob which passes its lowest point twice every second is (jamb 1995)

A. 0.25m

B. 0.45m

C. 0.58m

D. 1.00m [g=10ms^{-2}]

**Answer:** A

**Solution:** T = 2π√l/g; l = 2×3.142√l/10

L= 0.25m

8. A test-tube of radius 1.0cm is loaded to 8.8g. If it is placed upright in water, find the depth to which it would sink. (Jamb 2003)

A. 2.8cm

B. 5.2cm

C. 25.5cm

D. 28.0cm [g=10ms^{-2}, density of water = 1000kgm^{-3}]

**Answer:** A

**Solution:** mass of h_{2}o displaced = 8.8g

Volume of h_{2}o displaced = mass/ density = 8.8g/1g/cm

= 8.8cm^{3}; volume of cylinder = πr^{2}h = 8.8

22/7 ×1×1×h= 8.8;

H= 2.8cm

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