2 Marks Questions

Question 12 Marks

A one metre long stretched string having a mass of 40g is attached to a tuning fork. The fork vibrates at 128Hz in a direction perpendicular to the string. What should be the tension in the string if it is to vibrate in four loops?

Answer

A string of mass 40g is attached to the tuning fork

$\text{m}=(40\times10^{-3})\text{kg/m}$

The fork vibrates with $\text{f}=128\text{Hz}$

$\lambda=0.5\text{m}$

$\text{v}=\text{f}\lambda=128\times0.5=64\text{m/s}$

$\text{v}=\sqrt{\frac{\text{T}}{\text{m}}}$

$\Rightarrow\text{T}=\text{v}^2\text{m}=163.84\text{N}$

$\Rightarrow164\text{N}$

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Question 22 Marks

Figure shows a wave pulse at t = 0. The pulse moves to the right with a speed of 10cm/s. Sketch the shape of the string at t = 1s, 2s and 3s.

Answer

At t = 1 sec, s₁ = vt = 10 × 1 = 10cm
t = 2 sec, s₂ = vt = 10 × 2 = 20cm
t = 3 sec, s₃ = vt = 10 × 3 = 30cm

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Question 32 Marks

In the arrangement shown in figure, the string has a mass of 4.5g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley? Take g = 10m/s².

Answer

Total length of string $2+0.25=2.25\text{mt}$

Mass per unit length $\text{m}=\frac{4.5\times10^{-3}}{2.25}=2\times10^{-3}\text{kg/m}$

$\text{T}=2\text{g}=20\text{N}$

Wave speed,

$\text{v}=\sqrt{\Big(\frac{\text{T}}{\text{m}}\Big)}=\sqrt{\frac{20}{(2\times10^{-3})}}=\sqrt{10^4}=10^2\text{m/s}=100\text{m/s}$

Time taken to reach the pully, $\text{t}=\Big(\frac{\text{s}}{\text{v}}\Big)=\frac{2}{100}=0.02\text{sec}.$

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Question 42 Marks

A sonometer wire having a length of 1.50m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256Hz. What is the speed of the transverse wave on the wire?

Answer

First harmonic be f₀, second harmonic be f₁

$\therefore\text{f}_1=2\text{f}_0$

$\Rightarrow\text{f}_0=\frac{\text{f}_1}{2}$

$\text{f}_1=256\text{Hz}$

$\therefore\ $1^st harmonic or fundamental frequency

$\text{f}_0=\frac{\text{f}_1}{2}=\frac{256}{2}=128\text{Hz}$

$\frac{\lambda}{2}=1.5\text{m}\Rightarrow\lambda=3\text{m}$ (when fundamental wave is produced)

⇒ Wave speed = $\text{v}=\text{f}_0\text{Ql}=384\text{m/s}.$

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Question 52 Marks

The equation of a standing wave, produced on a string fixed at both ends, is

$\text{y}=(0.4\text{cm})\sin\big[(0.314\text{cm}^{-1})\text{x}\big]\cos\big[(600\pi\text{s}^{-1})\text{t}\big]$

What could be the smallest length of the string?

Answer

The equation of the standing wave is given by

$\text{y}=(0.4\text{cm})\sin\big[0.314\text{cm}^{-1}\text{x}\big]\cos\big[(6.00\pi\text{s}^{-1})\text{t}\big]$

$\Rightarrow\text{k}=0.314=\frac{\pi}{10}$

$\Rightarrow\frac{2\pi}{\lambda}=\frac{\pi}{10}$

$\Rightarrow\lambda=20\text{cm}$

for smallest length of the string, as wavelength remains constant, the string should vibrate in fundamental frequency

$\Rightarrow\text{l}=\frac{\lambda}{2}=\frac{20\text{cm}}{2} = 10\text{cm}$

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Question 62 Marks

The length of the wire shown in figure, between the pulleys is 1.5m and its mass is 12.0g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

Answer

$\text{l}=1.5\text{m},\ \text{mass}=12\text{g}$

$\Rightarrow\text{m}=\frac{12}{1.5}\text{gm}=8\times10^{-3}\text{kg/m}$

$\text{T}=9\times\text{g}=90\text{N}$

$\lambda=1.5\text{m},\ \text{f}_1=\frac{2}{\text{2l}}\sqrt{\frac{\text{T}}{\text{m}}}$

[for, second harmonic two loops are produced]

$\text{f}_1=2\text{f}_0$

$\Rightarrow70\text{Hz}$

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Question 72 Marks

A piano wire weighing 6.00g and having a length of, 90.0cm emits a fundamental frequency corresponding to the "Middle C" (v = 261.63Hz). Find the tension in the wire.

Answer

$\text{l}=90\text{cm}=0.9\text{m}$

$\text{m}=\Big(\frac{6}{90}\Big)\text{g/cm}=\Big(\frac{6}{900}\Big)\text{kg/mt}$

$\text{f}=261.63\text{Hz}$

$\text{f}=\frac{1}{2\text{l}}\sqrt{\Big(\frac{\text{T}}{\text{m}}\Big)}\Rightarrow\text{T}=1478.52\text{N}=1480\text{N}.$

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Question 82 Marks

A tuning fork of frequency 440Hz is attached to a long string of linear mass density 0.01kg/m kept under a tension of 49N. The fork produces transverse waves of amplitude 0.50mm on the string.

Find the wave speed and the wavelength of the waves.
Find the maximum speed and acceleration of a particle of the string.
At what average rate is the tuning fork transmitting energy to the string?

Answer

$\text{f}=440\text{Hz},\ \text{m}=0.01\text{kg/m},\text{T}=49\text{N},\text{r}=0.5\times10^{-3}\text{m}$

$\text{v}=\sqrt{\frac{\text{T}}{\text{m}}}=70\text{m/s}$
$\text{v}=\lambda\text{f}\Rightarrow\lambda=\frac{\text{v}}{\text{f}}=16\text{cm}$
$\text{P}_\text{avearage}=2\pi^2\text{mvr}^2\text{f}^2=0.67\text{W}.$

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Question 92 Marks

A particular guitar wire is 30.0cm long and vibrates at a frequency of 196Hz when no finger is placed on it. The next higher notes on the scale are 220Hz, 247Hz, 262Hz and 294Hz. How far from the end of the string must the finger be placed to play these notes?

Answer

$\text{l}_1=30\text{cm}=0.3\text{m}$

$\text{f}_1=196\text{Hz},\ \text{f}_2=220\text{Hz}$

We know $\text{f}\propto\Big(\frac{1}{\text{l}}\Big)$ (as V is constant for a medium)

$\Rightarrow\frac{\text{f}_1}{\text{f}_2}=\frac{\text{l}_2}{\text{l}_1}$

$\Rightarrow\text{l}_2=26.7\text{cm}$

Again $\text{f}_3=247\text{Hz}$

$\Rightarrow\frac{\text{f}_3}{\text{f}_1}=\frac{\text{l}_1}{\text{l}_3}$

$\Rightarrow\frac{0.3}{\text{l}_3}$

$\Rightarrow\text{l}_3=0.224\text{m}=22.4\text{cm}$ and $\text{l}_3=20\text{cm}$

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Question 102 Marks

A wire of length 2.00m is stretched to a tension of 160N. If the fundamental frequency of vibration is 100Hz, find its linear mass density.

Answer

$\text{l}=2\text{m},\ \text{f}_0=100\text{Hz},\ \text{T}=160\text{N}$

$\text{f}_0=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\text{m}}}$

$\Rightarrow\text{m}=1\text{g/m}.$ So, the linear mass density is 1g/m.

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Question 112 Marks

The displacement of the particle at x = 0 of a stretched string carrying a wave in the positive x-direction is given by $\text{f(t)}=\text{a}\sin\Big(\frac{\text{t}}{\text{T}}\Big).$ The wave speed is v. Write the wave equation.

Answer

At x = 0, $\text{f(t)}=\text{a}\sin\Big(\frac{\text{t}}{\text{T}}\Big).$

Wave speed = v

$\Rightarrow\lambda$ = wavelength = vT (T = Time period)

So, general equation of wave

$\text{Y}=\text{A}\sin\bigg[\Big(\frac{\text{t}}{\text{T}}\Big)-\Big(\frac{\text{x}}{\text{vT}}\Big)\bigg]$ $\Bigg[$ because $\text{y}=\text{f}\bigg(\Big(\frac{\text{t}}{\text{T}}\Big)-\Big(\frac{\text{x}}{\lambda}\Big)\bigg)\Bigg]$

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Question 122 Marks

A wire, fixed at both ends is seen to vibrate at a resonant frequency of 240Hz and also at 320Hz.

What could be the maximum value of the fundamental frequency?
If transverse waves can travel on this string at a speed of 40m/s, what is its length?

Answer

This wire makes a resonant frequency of 240Hz and 320Hz.

The fundamental frequency of the wire must be divisible by both 240Hz and 320Hz.

So, the maximum value of fundamental frequency is 80Hz.
Wave speed, v = 40m/s

$\Rightarrow80=\frac{1}{2\text{l}}\times40$

$\Rightarrow0.25\text{m}.$

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Question 132 Marks

Three resonant frequencies of a string are 90, 150 and 210Hz.

Find the highest possible fundamental frequency of vibration of this string.
Which harmonics of the fundamental are the given frequencies?
Which overtones are these frequencies.
If the length of the string is 80cm, what would be the speed of a transverse wave on this string?

Answer

The resonant frequencies of a string are

f₁ = 90Hz, f₂ = 150Hz, f₃ = 120Hz

The highest possible fundamental frequency of the string is f = 30Hz

[because f₁, f₂ and f₃ are integral multiple of 30Hz]

The frequencies are f₁ = 3f, f₂ = 5f, f₃ = 7f

So, f₁, f₂ and f₃ are 3rd harmonic, 5^th harmonic and 7^th harmonic respectively.

The frequencies in the string are f, 2f, 3f, 4f, 5f, .....

So, 3f = 2^nd overtone and 3^rd harmonic

5f = 4^th overtone and 5^th harmonic

7f = 6^th overtone and 7^th harmonic

Length of the string is $\text{l}=80\text{cm}$

$\Rightarrow\text{f}_1=\Big(\frac{3}{2\text{l}}\Big)\text{v}$ (v = velocity of the wave)

$\Rightarrow90=\Big\{\frac{3}{(2\times80)}\Big\}\times\text{K}$

$\Rightarrow\text{K}=\frac{(90\times2\times80)}{3}$

$=4800\text{cm/s}$

$=48\text{m/s}.$

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Question 142 Marks

A pulse travelling on a string is represented by the function:

$\text{y}=\frac{\text{a}^3}{(\text{x}-\text{vt})^2+\text{a}^2},$

where a = 5mm and v = 20cm/s. Sketch the shape of the string at t = 0, 1s and 2s.Take x = 0 in the middle of the string.

Answer

The pulse is given by, $ \text{y}=\Bigg[\frac{(\text{a}^3)}{\big\{(\text{x}-\text{vt})^2+\text{a}^2\big\}}\Bigg]$
a = 5mm = 0.5cm, v = 20cm/s
At t = 0s, $\text{y}=\frac{\text{a}^3}{(\text{x}^2+\text{a}^2)}$
The graph between y and x can be plotted by taking different values of x.
(left as exercise for the student)
Similarly, at t = 1s, $\text{y}=\frac{\text{a}^3}{\big\{(\text{x}-\text{v})^2+\text{a}^2\big\}}$
and at t = 2s, $\text{y}=\frac{\text{a}^3}{\big\{(\text{x}-2\text{v})^2+\text{a}^2\big\}}$

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Question 152 Marks

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6cm. Find the length of the string.

Answer

Let there be ‘n’ loops in the 1^st case

⇒ length of the wire, $\text{l}=\Big(\frac{\text{n}\lambda_1}{2}\Big)$ $[\lambda_1=2\times2=4\text{cm}]$

So there are (n + 1) loops with the 2^nd case

⇒ length of the wire, $\text{l}=\Big\{\frac{(\text{n}+1)\lambda_2}{2}\Big\}$ $[\lambda=2\times1.6=3.2\text{cm}]$

$\Rightarrow\frac{\text{n}\lambda_1}{2}=\frac{(\text{n}+1)\lambda_2}{2}$

$\Rightarrow\text{n}\times4=(\text{n}+1)(3.2)$

$\Rightarrow\text{n}=4$

$\therefore\ $length of the string, $\text{l}=\frac{\text{n}\lambda_1}{2}=8\text{cm}.$

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Question 162 Marks

A steel wire fixed at both ends has a fundamental frequency of 200Hz. A person can hear sound of maximum frequency 14kHz. What is the highest harmonic that can be played on this string which is audible to the person?

Answer

Fundamental frequency $\text{f}_1=200\text{Hz}$

Let l₄Hz be nth harmonic

$\Rightarrow\frac{\text{F}_2}{\text{F}_1}=\frac{14000}{200}$

$\Rightarrow\frac{\text{NF}_1}{\text{F}_1}=70$

$\Rightarrow\text{N}=70$

$\therefore\ $The highest harmonic audible is 70^th harmonic

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Question 172 Marks

A 200Hz wave with amplitude 1mm travels on a long string of linear mass density 6g/m kept under a tension of 60N.

Find the average power transmitted across a given point on the string.
Find the total energy associated with the wave in a 2.0m long portion of the sring.

Answer

$\text{A}=1\text{mm}=10^{-3}\text{m},\ \text{m}=6\text{g/m}=6\times10^{-3}\text{kg/m}$

$\text{T}=60\text{N},\text{f}=200\text{Hz}$

$\therefore\text{v}=\sqrt{\frac{\text{T}}{\text{m}}}=100\text{m/s}$

$\text{P}_\text{average}=2\pi^2\text{mv}\text{A}^2\text{f}^2=0.47\text{W}$
Length of the string is 2 m. So, $\text{t}=\frac{2}{100}=0.02\text{sec}.$

Energy $=2\pi^2\text{mvf}^2\text{A}^2\text{t}=9.46\text{mJ}.$

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