KPK G10 Physics Class 10 Notes Chapter 10 (Simple Harmonic Motion and Waves) physics notes for class 10. These include short questions, Conceptual Questions, Comprehensive Questions, and numerical Problems.
Table of Contents
Physics Class 10 Notes Chapter 10 Conceptual Questions
Q.1) Is every oscillatory motion simple harmonic? Give examples.
Answer:
Not all vibratory or oscillatory motions are simple harmonic motion since all restoring forces are not proportional to the displacement. Any restoring force can cause oscillatory motion.
Example:
An electrocardiogram traces the periodic pattern of a beating heart, but the motion of the recorder needle is not a simple harmonic motion. As the restoring force, in this case, is not always proportional to the displacement from the equilibrium position.
Q.2) For a particle with simple harmonic motion, at what point of the motion does the velocity attain maximum magnitude? Minimum magnitude?
Answer:
Maximum velocity:
For a particle with simple harmonic motion, the magnitude of velocity is maximum at the mean position or equilibrium position.
Minimum velocity:
The magnitude of velocity is minimum at extreme positions.
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Q.3) Is the restoring force on a mass attached to spring in simple harmonic motion ever zero? If so, where?
Answer:
Yes, the restoring force on a mass attached to spring in simple harmonic motion is zero at mean position or equilibrium position. i.e
We know from Hook’s law that the restoring force acting on the mass attached to the spring is:
Fre = -kx
here, x is the displacement from the mean position. At mean position x = 0 so, the restoring force is zero at the mean position.
Q.4) If we shorten the string of a pendulum to half its original length, what is the affect on its time period and frequency?
Answer:
The time period of simple pendulum is given by,
Hence we can see from above relations that if the length of the string of the pendulum is shortened to half, the time period of the pendulum will decrease by a factor 1/√2 and the frequency is increased by the factor √12.
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Q.5) A thin rope hangs from dark high tower so that its upper end is not visible. How can the length of rope be determined?
Answer:
If we hang a small mass at the end of the rope then it will become a simple pendulum. Let’s suppose that the length “l” of the pendulum is equal to the length of the rope. Now, the time period of a simple pendulum is,
From the second equation if we know the value of the time period of the pendulum then we can find the length of the rope.
Q.6) Suppose you stand on a swing instead of sitting on it. Will your frequency of oscillation increase or decrease?
Answer:
The time period of the swing having length “l” from the pivot point the center of mass of the swinger is given by.
Q.7) Explain the difference between the speed of a transverse wave traveling along a cord and the speed of a tiny colored part of the cord.
Answer:
The speed of the transverse wave is along the cord in the direction of the motion of the wave while the direction of the speed of the tiny colored part of the cord is perpendicular to the direction of the motion of the wave.
Q.8) Why wave refract at the boundary of shallow and deep water?
Answer:
When a wave travels from shallow to deep or deep to shallow water its velocity changes, this change in velocity causes refraction of the wave.
Q.9) What is the effect on diffraction if the opening is made small?
Answer:
If the opening is made small or equal to the wavelength of the wave, then the wave spread out in all directions and turn into circular waves. While, if the opening is large, then the wave will only diffract at the edges of an opening and the central part will remain unaffected. As shown in the figure.
What is the effect of diffraction if the opening is made small?
If the opening is made small or equal to the wavelength of the wave, then the wave spread out in all directions and turns into circular waves. While, if the opening is large, then the wave will only diffract at the edges of an opening and the central part will remain unaffected. As shown in the figure.
Physics Class 10 Notes Comprehensive Questions Chapter 10
Q.1) What is simple Harmonic Motion (SHM)? What are the conditions for an object to oscillate with SHM?
Answer:
Simple Harmonic Motion (SHM):
“To and fro motion of an object about its mean position in equal intervals of time is known as a simple harmonic motion”
Conditions for Simple Harmonic Motion:
Conditions for simple harmonic motion are:
- An object should vibrate or oscillate about its mean position.
- The acceleration produced due to restoring force should always direct towards the mean position.
- The acceleration should always be directly proportional to the displacement from the mean position.
Mathematically.
a ∝-x
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Q.2) Show that the mass spring system executes simple harmonic motion (SHM).
Answer:
Motion of Mass Attached to spring:
Consider a mass attached to a spring and is placed on a smooth, friction-less surface such that one end of the spring is attached to a solid support and other end to the mass. When the mass is at rest i.e. at its mean position ‘O’ then the whole system is said to be in equilibrium as shown in fig (a).
Now if we apply some external force Fext on the spring such that the spring gets stretched by pulling the mass outwards then the length of the spring will increase. Let this increase in length is ‘x’ due to the displacement of the mass from point ‘O’ to point ‘A’ as shown in fig (b).
According to Hook’s law, force applied will be directly proportional to the displacement i.e. F α x or F = kx
Here ‘k’ is the constant of proportionality known as the spring constant.
Now if we remove the force, the mass will move towards its mean position but due to inertia it will not stay there but go further in the same direction such that there is a compression in the string and in this way the mass will keep on moving to and fro about its mean position. That motion of the mass towards the mean position is due to restoring force Fres. This force is given by,
Fres = -kx
(1)
Negative sign here is due to the restoring force.
Such type of motion is known as simple harmonic motion.
Now according to Newton’s law, we have,
F = ma
Comparing this with equation (1), we get
ma = -kx ⇒ a = – (k/m) x
But as (k/m) is a constant so we can say that acceleration and the displacement are directly proportional to each other, i.e.
a ∝-x
Hence, the mass spring system executes simple harmonic motion (SHM)
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Q.3) What is simple pendulum? Diagrammatically show the forces acting on simple pendulum. Also show that simple pendulum executes simple harmonic motion.
Answer:
Simple Pendulum:
Consider a small bob of mass ‘m’ is suspended by a string of length ‘l’ from friction-less fixed support as shown in the figure. This arrangement is known as a simple pendulum.
Forces act on a simple pendulum:
A simple pendulum is driven by the force of gravity due to the weight of the suspended mass. (i.e. W = mg). The weight of the simple pendulum is resolved into two components mgcosθ and mgsinθ. The mgcosθ acts in opposition to the tension in the string T. While the mgsinθ provides restoring force to the pendulum, as shown in the figure below.
Simple pendulum as SHM:
Consider △QRS from the above figure. The restoring force is given by
Fres = -mgsinθ ………………………………. (1)
From the figure, for a small angle, the arc length ‘s’ is nearly the same as displacement ‘x’. Therefore,
sinθ = x/l ………………………………………..(2)
Putting equation (2) into (1). we get
Fres = -mg(x/l) ……………………………… (3)
From Newtons 3rd law of motion, this restoring force is equal to,
Fres = ma ………………………………………. (4)
Comparing equation 3 and (4)
ma =-mg (x/l)
or
a = -x(g/l) …………………………………….. (5)
Since for a fixed length of the pendulum and constant value of ‘g’, g/l is a constant quantity. Hence we can write equation (5) as
a ∝-x
Hence acceleration is directly proportional to the displacement, So, simple pendulum executes the simple harmonic motion.
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Q.4) What is wave motion? How waves can be categorized?
Answer:
Wave motion:
Wave motion is related to oscillation when the energy moves through the wave the particles of the medium executes simple harmonic motion about their equilibrium position.
Example:
Dropping a stone in a pond water produces ripples on the surface of water. The particles of water that absorb energy and start oscillating from the impact of the stone.
Categorization of waves:
The waves are mainly catagorized into two types of waves.
- Mechanical waves:
The waves produced by the oscillation of material particles are called mechanical waves. Examples of mechanical waves are sound waves, seismic waves etc. - Electromagnetic waves:
The waves that propagate by the oscillation of electric and magnetic field are called electromagnetic waves. These waves do not require and medium to propagate. These are a combination of traveling electric and magnetic field. Examples x-rays, microwaves, radio waves etc.
Waves are also classified as transverse and longitudinal on the basis of there propagation and direction of distribution.
Transverse Waves:
‘The wave in which the vibrations of the individual particles of the medium is perpendicular to the direction of propagation of waves are called transverse waves”
Example:
If we take a string and fix its one end to a solid support. Now if we make a jerk to the string by moving it up and down then the waves produced in the string are transverse waves.
Longitudinal Waves:
“The waves in which the particles of the medium vibrate about their mean position parallel to the direction of propagation of waves are called longitudinal waves”
Example:
Sound waves are an example of longitudinal waves as they are produced due to the displacement of the air molecules forward and backward along the same axis.
Q.5) How wave transport energy without carrying the material medium? Also, describe the connection between oscillatory motion and waves.
Answer:
Wave motion is related to oscillation when the energy moves through the wave the particles of the medium executes simple harmonic motion about their equilibrium position.
Example:
Dropping a stone in a pond water produces ripples on the surface of water. The particles of water that absorb energy and start oscillating from the impact of the stone. These particles transfer some of its energy to the neighboring particles which also start vibrating. In this way gradually other particles on the water surface also start oscillating and energy is spread out through the water pond with out carrying the material medium.
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Q.7) Using ripple tank explain reflection, refraction and diffraction of waves.
Answer:
Properties of Waves:
Consider a ripple tank such that it has a base of the glass. In order to illuminate the tank, x we fall light from the above, so that the light may shine through the water as shown in figure (a). By doing so, crusts in waves will appear by bright-line and troughs will be by dark lines.
1) Reflection of Water:
It is the nature of waves that they are reflected upon hitting a rigid barrier such that the angle of reflected waves will be same as that of coming waves. Speed of coming and reflected waves will also be same. Therefore in a ripple tank, the reflection process of waves can be visualized by placing an upright barrier in water as shown in figure (b) and (c).
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2) Refraction of Waves:
Waves moving in different mediums will have different speeds. The speed of any wave will be greater if the medium is rare and will be slower if the medium will be denser. Therefore when waves enter from a rare medium into a denser medium then they bend a little from their actual path. This bending of waves is called the refraction of waves.
In order to understand this phenomenon on a ripple tank, let place a plastic sheet at the bottom of the tank. By doing so we can see the refraction of waves at the edge of the medium. Also it is observed that the speed of the waves in deeper water is greater than the shallow water as shown in figure (d).
3) Diffraction of Waves:
In order to understand this phenomenon, produce the straight waves in a ripple tank and then place two obstacles in the way of these waves. The separation between these two obstacles should be comparable to the wavelength of the coming waves.
When the waves will pass through the slit between these two obstacles then they will bend circularly around the slit as shown in figure (e). This bending of waves is known as diffraction.
Diffraction will be maximum if the separation between obstacles will be exactly equal to the wavelength of the waves.
Numerical Questions, numerical problems in physics class 10 Simple Harmonic Motion and Waves Chapter 10
Q.1) A mass hung from a spring vibrates 15 times in 12 s. Calculate (a) the frequency and (b) the period of the vibration.
Answer:
Given Data:
Time = t = 12 s
No. of cycles = 15 cycles
To find:
a). Frequency = f = ?
b). Period of vibration = T = ?
Calculation:
a).
f = No. of cycles/time
f = 15/12
f = 1.25 Hz
b).
We know that,
T = 1/f
T = 1/1.25
T = 0.8 s
Q.2) A spring requires a force of 100.0 N to compress it to a displacement of 4 cm. What is its spring constant?
Answer:
Given Data:
Force = F = 100 N
displacement = x = 4 cm = 0.04 m
To find:
Spring constant = k =?
Calculation:
From Hook’s law, we know that
F = kx
or
k = F/x
k = (100/0.04)
k = 2500 Nm-1
Q.4) Calculate the period and frequency of a propeller on a plane if it completes 250 cycles in 5.0 s.
Answer:
Given Data:
Time = t = 5s
No. of cycles = 250 cycles
To find:
a). Frequency = f = ?
b). Period = T = ?
Calculation:
a).
f = No. of cycles/time
f = 250/5
f = 50 Hz
b).
We know that,
T = 1/f
T = 1/ 50
T = 0.02 s
Q.5) Water waves with wavelength 2.8 m, produced in a ripple tank, travel with a speed of 3.80 m/s. What is the frequency of the straight vibrator that produced them?
Answer:
Given data:
Wavelength = λ = 2.8 m
speed = v = 3.80 m/s
To find:
frequency = f = ?
Calculation:
Universal wave equation is,
v = f λ
or
f = v/ λ
f = 3.80/2.8
f = 1.35 Hz.
Q.6) The distance between successive crests in a series of water waves is 4.0 m, and crest travel 9.0 m in 4.5 s. What is the frequency of the waves?
Answer:
Given data:
Wavelength = distance between successive crests = λ =4.0 m
Distance travelled = d = 9 m
time take = t =4.5 s
To find:
frequency = f =?
Calculation:
First, lets find the wavespeed, i.e.
v = distance travelled/time take
v = d/t
v = 9/4.5
v = 2 m/s
Now, the universal wave equation is,
v = f λ
or
f = v/λ
f = 2/4
f = 0.5 Hz.
Q.7) A station broadcasts an AM radio wave whose frequency is 1230 x 103 Hz (1230 kHz on the dial). Find the distance between adjacent crests in each wave.
Answer:
Given data:
Frequency = f = 1230 kHz = 1230 × 103 Hz
speed of radio wave = speed of light = c = 3 × 108 m/s
To find:
Distance between adhacnet crests = wavelength = λ = ?
Calculation:
Universal wave equation is,
v = f λ
or
λ = v/f
λ = (3 × 108)/(1230 × 103 )
λ = 243 m.
Numerical Problem No. 3 is missing.