A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does he disturbance take to reach the other end ?

M = 2.50 kg T = 200 N l = 20.0 mMass per unit length, \\(\mu=M/l = 2.50/20=0.125Kg\, m^{-1}\\) The velocity (v) of the transverse wave in the string is given by the relation:\\( v = \sqrt{T / \mu}\\)= \\(\sqrt{200/0.125} = \sqrt{1600} = 40\ m/s\\)∴ Time taken by the disturbance to reach the other end,\\(t = l/v = 20/40 = 0.5 s\\)

A string is stretched between fixed points separated by \\(75\ cm\\). It is observed to have resonant frequencies of \\(420\ Hz\\) and \\(315\ Hz\\). There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Let \\(315Hz\\) be the \\(nth\\) harmonic of the string.Thus \\(\dfrac{nc}{2l}=315Hz\\)and \\(\dfrac{(n+1)c}{2l}=420Hz\\)\\(\implies \dfrac{n+1}{n}=\dfrac{4}{3}\\)\\(\implies n=3\\)Thus lowest resonant frequency=\\(\dfrac{c}{2l}=\dfrac{315}{n}Hz=105Hz\\)

A uniform rope of length \\(L\\) and mass \\({m}_{1}\\) hangs vertically from a rigid support. A block of mass \\({m}_{2}\\) is attached to the free end of the rope. A transverse pulse of wavelength \\({\lambda}_{1}\\) is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is \\({\lambda}_{2}\\). The ratio \\({{\lambda}_{2}}/{{\lambda}_{1}}\\) is:

\\(T_1=m_2g\\)\\(T_2=(m_1+m_2)g\\)Velocity \\(\propto \sqrt{T}\\) \\(\Rightarrow \lambda \propto \sqrt{T}\\)\\(\Rightarrow \dfrac{\lambda_1}{\lambda_2}=\dfrac{\sqrt{T_1}}{\sqrt{T_2}}\\)\\(\Rightarrow \dfrac{\lambda_2}{\lambda_1}=\sqrt{\dfrac{m_1+m_2}{m_2}}\\)

For a stationary wave \\(y=4\sin\left(\displaystyle\frac{\pi x}{15}\right)\cos(96\pi t)\\), the distance between a node and the next antinodes is.

Comparing given equation with standard equation \\(y=\displaystyle 2a\sin \frac{2\pi x}{\lambda}\cos \displaystyle\frac{2\pi vt}{\lambda}\\)Given us \\(\displaystyle\frac{2\pi}{\lambda}=\frac{\pi}{15}\\)\\(\Rightarrow \lambda=30\\)Distance between nearest node and antinode \\(=\displaystyle\frac{\lambda}{4}=\frac{30}{4}=7.5\\)

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is \\(3.5 \times 10^{-2}\ kg \\) and its linear mass density is \\(4.0 \times 10^{-2}\ kg \ m^{-1}\\). What is (a) the speed of a transverse wave on the string, and (b) the tension in the string ?

(a) Mass of the wire, \\(m=3.5\times 10^{-2}kg\\)Linear mass density, \\(\mu=m/l = 4.0\times 10^{-2}kg\,m^{-1}\\)Frequency of vibration, \\(f=45Hz\\)\\(\therefore\\) Length of the wire, \\(l=\displaystyle \frac{m}{\mu} = \frac{3.5\times 10^{-2}}{4.0\times 10^{-2}}=0.875m\\)The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:\\(\lambda =2l/n\\)where,\\(n=\\) Number of nodes in the wireFor fundamental node, \\(n=1\\):\\(\lambda=2l\\)\\(\lambda =2\times 0.875=1.75m\\)The speed of the transverse wave in the string is given as:\\(v=f\lambda=45\times 1.75=78.75m/s\\)(b) The tension produced in the string is given by the relation:\\(T=v^2\mu\\)\\(=(78.75)^2\times 4.0\times 10^{-2}=248.06\,N\\)

A standing wave having \\(3\\) nodes and \\(2\\) antinodes is formed between two atoms having a distance \\(1.21\mathring { A } \\) between them. The wavelength of the standing wave is

The given situation can be shown asGiven the distance between the two atoms \\(=1.21\mathring { A } \\)so, the wavelength of the standing wave is\\(\cfrac { \lambda }{ 2 } +\cfrac { \lambda }{ 2 } =\lambda =1.21\mathring { A } \\)

A tuning fork of frequency \\(340 Hz\\) is vibrated just above the tube of \\(120 cm\\) height. Water is poured slowly in the tube. What is the minimum height of water necessary for the resonance? \\((\\)speed of sound in air \\(= 340 m/s)\\)

We have \\(v=v\lambda\\)or \\(\displaystyle \lambda = \frac {v}{v} = \frac {340 m/s}{340 Hz}= 1m\\)First resonating length,\\(\displaystyle l_1=\frac {\lambda }{4} = \frac {1}{4}m=25 cm\\)Second resonating lenght,\\(\displaystyle l_2=\frac {3\lambda }{4} = \frac {3\times 1m}{4}=75 cm\\)Third resonating lenght,\\(\displaystyle l_3=\frac {5\lambda }{4} = \frac {5\times 1m}{4}=125 cm\\)So third resonance is not possible since the length of the tube is \\(120 cm\\)\\(\therefore \\) Minimum height of water necessary for resonance \\(=120-75=45 cm\\)

Show that all harmonics are present in case of an air column vibrating in a pipe open at both ends.

For a tube open at both ends the antinodes should be formed at the open ends as shown below: [Ref. image] In the fundamental mode or case 1 as shown above There are one node at the center and two antinodes at the open ends.We can see that \\(\dfrac{\lambda_1}{2} = L\\), where \\(L\\) is the length of pipe.Let the velocity of the sound wave be '\\(v\\)'Then frequency \\(\nu_1 = \dfrac{v}{\lambda_1} = \dfrac{v}{2L}\\)For second case similarly we see\\(\lambda_2 = L\\)So, \\(\nu_2 = \dfrac{v}{\lambda_2} = \dfrac{v}{L} = \dfrac{2v}{2L}\\)For third case\\(\dfrac{3\lambda_3}{2}=L\\)So, \\(\nu_3 = \dfrac{v}{\lambda_3} = \dfrac{3v}{2L}\\)Similarly if we take for \\(n^{th}\\) harmonic we will get\\(\nu_n = \dfrac{v}{\lambda_n} = \dfrac{nv}{2L}\\)We can see from above that all the harmonics\\(\dfrac{v}{2L} , \dfrac{2v}{2L}, \dfrac{3v}{2L}.......\dfrac{nv}{2L}\\)is present in case of air column vibrating in a pipe open at both ends and it can be explained because '\\(n\\)' can take any positive integer value from \\(1\\) to infinity.

A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound. (b) the transmitted sound? Speed of sound in air is \\(340\ ms^{-1} \\) and in water \\(1486 \ ms^{-1}\\).

(a) Frequency of the ultrasonic sound, \\(f= 1000 kHz=10^6Hz\\)Speed of sound in air, \\(V_a = 340 m/s\\)The wavelength \\((\lambda_r)\\) of the reflected sound is given by the relation:\\(\lambda_r = V/f\\)\\(=340/10^6 = 3.4\times 10^{-4}m\\).(b) Frequency of the ultrasonic sound, \\(f=1000 kHz = 10^6 Hz\\)Speed of sound in water, \\(V_w = 1486\ m/s\\)The wavelength of the transmitted sound is given as:\\(\lambda_r=1486 / 10^6 = 1.49\times 10^{-3}m\\).

A wave in a string has an amplitude of \\(2 cm\\). The wave travels in the +ve direction of x-axis with a speed of \\(128ms^{-1}\\) and it is noted that \\(5 \\)complete waves fit in 4m length of the string. The equation describing the wave is

Amplitude \\(=2\,cm=0.02\,m\\)\\(v=128\,m/s\\)\\(5\lambda=4 \implies \lambda=\dfrac 45 =0.8\,m\\)\\(\nu=\dfrac{128}{0.8}=160\,Hz\\)\\(\omega=2\pi \nu=2\pi \times 160=1005\\)\\(k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{0.8}=7.85\\)Therefore,\\(y=0.02\,sin(7.85x-1005t)\\)

Reflection of Waves | Definition, Examples, Diagrams

definition

Incident, Reflected and Transmitted waves at a boundary

If a pulse is introduced at the left end of the rope, it will travel through the rope towards the right end of the medium. This pulse is called the incident pulse since it is incident towards (i.e., approaching) the boundary with the pole. When the incident pulse reaches the boundary, two things occur:
1. A portion of the energy carried by the pulse is reflected and returns towards the left end of the rope. The disturbance that returns to the left after bouncing off the pole is known as the reflected pulse.

2. A portion of the energy carried by the pulse is transmitted to the pole, causing the pole to vibrate.

example

Write the equation of reflected wave for a given incident wave at a rigid boundary

Example: Find the phase change between incident and reflected sound wave from a fixed wall by writing equations of incident and reflected wave.

Solution:
When a wave is reflected from a wall, the wave is inverted . As the phase difference between a wave and its inverted wave is

\overset{n}{ˉ}

, the phase difference is

π

y = A c o s (k x + ω t)

y_{r} = - A c o s (k x + ω t)

= A c o s (k x + ω t + π)

\Rightarrow ϕ = π

example

Reflection of sound

Sound waves (like any other wave) return back in the same medium obeying the laws of reflection. The only requirement for the reflection of the sound wave is that the reflecting surface must be bigger than the wavelength of the wave. Reflection of sound wave is seen in echoes, megaphone, ear trumpet etc.

definition

Explain refraction of sound

Refraction of sound waves is most evident in situations in which the sound wave passes through a medium with gradually varying properties. For example, sound waves are known to refract when traveling over water. Even though the sound wave is not exactly changing media, it is traveling through a medium with varying properties; thus, the wave will encounter refraction and change its direction.

result

Equation of reflected wave for a given incident sound wave

Example: The displacement of the medium in a sound wave is given by the equation;

y_{1} = A c o s (a x + b t)

where a & b are positive constants.
The wave is reflected by an obstacle situated at x = 0 . The intensity of the reflected wave is 0.64 times that if the incident wave. The equation for the reflected wave is given as

y_{2} = \pm \frac{s}{1 0} A c o s (a x - b t)

Find

s

Solution:
equation for sound intensity is

I = \frac{1}{2} ρ ω^{2} A^{2} v

I_{r} = 0.64 I_{i}

\frac{1}{2} ρ ω^{2} A_{r}^{2} v = \frac{0 . 6 4}{2} ρ ω^{2} A_{i}^{2} v

A_{r}^{2} = 0.64 A_{i}^{2}

A_{r} = 0.8 A_{i} = 0.8 A

\frac{s}{1 0} A = 0.8 A

s = 8

example

Reflection of a given incident sound wave in an open tube

Example: A train of sound waves is propagated along a wide pipe and it is
reflected from an open end. If the amplitude of the waves is 0.002 cm, the frequency 1000 Hz and the wavelength 40 cm, Find the amplitude of vibration at a point 10 cm from open end inside the pipe.

Solution:

Wavelength is 40cm
Amplitude to be found at 10 cm

= \frac{4 0}{4} c m = \frac{λ}{4}

In open pipe from open pipe at a distance of

^{λ} /_{4}

always a node is formed. So, amplitude at 10 cm from open end will be zero because amplitude at node is zero.

example

Incident, reflected and transmitted sound waves at a boundary

The simplest situation of reflection and transmission occurs when waves are impinging normal to the surface. The case of a longitudinal wave incident on the interface between two media is shown in the above figure.

law

Laws of reflection of sound

Law 1 : Incident wave,normal to the reflecting surface and reflected wave at the point of incidence all lie in the same plane.
Law 2: Angle of incidence is equal to the angle of reflection.

Reflection of Waves

definition

Incident, Reflected and Transmitted waves at a boundary

example

Write the equation of reflected wave for a given incident wave at a rigid boundary

example

Reflection of sound

definition

Explain refraction of sound

result

Equation of reflected wave for a given incident sound wave

example

Reflection of a given incident sound wave in an open tube

example

Incident, reflected and transmitted sound waves at a boundary

law

Laws of reflection of sound

REVISE WITH CONCEPTS

Pulse Travelling Along a String With a Fixed/Open End

Standing Waves and Normal Modes

Quick Summary With Stories

Reflection Of Sound

A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does he disturbance take to reach the other end ?

A string is stretched between fixed points separated by $75 c m$ . It is observed to have resonant frequencies of $420 H z$ and $315 H z$ . There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

For a stationary wave $y = 4 sin (\frac{π x}{1 5}) cos (96 π t)$ , the distance between a node and the next antinodes is.

A standing wave having $3$ nodes and $2$ antinodes is formed between two atoms having a distance $1.21 \overset{˚}{A}$ between them. The wavelength of the standing wave is

A tuning fork of frequency $340 H z$ is vibrated just above the tube of $120 c m$ height. Water is poured slowly in the tube. What is the minimum height of water necessary for the resonance? $($ speed of sound in air $= 340 m / s)$

Show that all harmonics are present in case of an air column vibrating in a pipe open at both ends.

A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound. (b) the transmitted sound? Speed of sound in air is $340 m s^{- 1}$ and in water $1486 m s^{- 1}$ .

A wave in a string has an amplitude of $2 c m$ . The wave travels in the +ve direction of x-axis with a speed of $128 m s^{- 1}$ and it is noted that $5$ complete waves fit in 4m length of the string. The equation describing the wave is

definition

Incident, Reflected and Transmitted waves at a boundary

example

Write the equation of reflected wave for a given incident wave at a rigid boundary

example

Reflection of sound

definition

Explain refraction of sound

result

Equation of reflected wave for a given incident sound wave

example

Reflection of a given incident sound wave in an open tube

example

Incident, reflected and transmitted sound waves at a boundary

law

Laws of reflection of sound

REVISE WITH CONCEPTS

Pulse Travelling Along a String With a Fixed/Open End

Standing Waves and Normal Modes

Quick Summary With Stories

Reflection Of Sound

A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does he disturbance take to reach the other end ?

A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

For a stationary wave y=4sin(15πx​)cos(96πt), the distance between a node and the next antinodes is.

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5×10−2 kg and its linear mass density is 4.0×10−2 kg m−1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string ?

A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance 1.21A˚ between them. The wavelength of the standing wave is

A tuning fork of frequency 340Hz is vibrated just above the tube of 120cm height. Water is poured slowly in the tube. What is the minimum height of water necessary for the resonance? (speed of sound in air =340m/s)

Show that all harmonics are present in case of an air column vibrating in a pipe open at both ends.

A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound. (b) the transmitted sound? Speed of sound in air is 340 ms−1 and in water 1486 ms−1.

A wave in a string has an amplitude of 2cm. The wave travels in the +ve direction of x-axis with a speed of 128ms−1 and it is noted that 5complete waves fit in 4m length of the string. The equation describing the wave is

A string is stretched between fixed points separated by $75 c m$ . It is observed to have resonant frequencies of $420 H z$ and $315 H z$ . There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

For a stationary wave $y = 4 sin (\frac{π x}{1 5}) cos (96 π t)$ , the distance between a node and the next antinodes is.

A standing wave having $3$ nodes and $2$ antinodes is formed between two atoms having a distance $1.21 \overset{˚}{A}$ between them. The wavelength of the standing wave is

A tuning fork of frequency $340 H z$ is vibrated just above the tube of $120 c m$ height. Water is poured slowly in the tube. What is the minimum height of water necessary for the resonance? $($ speed of sound in air $= 340 m / s)$

A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound. (b) the transmitted sound? Speed of sound in air is $340 m s^{- 1}$ and in water $1486 m s^{- 1}$ .

A wave in a string has an amplitude of $2 c m$ . The wave travels in the +ve direction of x-axis with a speed of $128 m s^{- 1}$ and it is noted that $5$ complete waves fit in 4m length of the string. The equation describing the wave is