Wave Speed on a Stretched String MCQ Quiz in বাংলা - Objective Question with Answer for Wave Speed on a Stretched String - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Mar 15, 2025
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Wave Speed on a Stretched String Question 1:
The wave velocity in a guitar string of length l and radius r, made of a material of density ρ, is v. If another string of same length and radius, but made of a material of density 3ρ, is used, the wave velocity is nv, under the same tension. The value of n is
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 1 Detailed Solution
CONCEPT:
The wave velocity of a string under Tension T and mass per unit length μ
⇒ v = \(\sqrt{\frac{T}{μ} } \)
EXPLANATION:
For the first string,
Length = l, radius = r, density = ρ
The volume of the wire = πr2L
∴ density = mass/volume
⇒ volume = mass/density = m1/ρ
So, we have, πr2L = m1/ρ
⇒ m1/L = πr2ρ = mass per unit length = μ
So, velocity v = \(\sqrt{\frac{T}{\pi r^2\rho} } \) ---- (1)
Similarly for the second material of mass m2
⇒ m2/L = πr2(3ρ)
So, velocity v' = nv = \(\sqrt{\frac{T}{\pi r^ 2 3\rho} } = \frac{1}{\sqrt{3}} v = 0.58 v\)
∴ n = 0.58
Hence the correct answer is option 1.
Wave Speed on a Stretched String Question 2:
If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency 'n' of the string is given by following :
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 2 Detailed Solution
Concept:
Fundamental frequency:
The lowest frequency of any vibrating object is called a fundamental frequency.
The fundamental frequency of a string is given by
\(n = \frac{1}{{2l}}\;\sqrt {\frac{T}{m}}\;\)
Where l = length of the string, T = tension on the string and m = linear mass density
Calculation:
Given:
The fundamental frequency of a string (l) is given by:
\(⇒ n = \frac{1}{{2l}}\;\sqrt {\frac{T}{m}} \)
As T and m is constant
\(\therefore n\propto \frac {1}{l}\)
⇒ n1l1 = n2l2 = n3l3 = k [Where k = constant]
\(⇒ l_1=\frac{k}{n_1},\,\,\, l_2=\frac{k}{n_2},\,\,\, l_3=\frac{k}{n_3}\)
The original length of the string is
\(\Rightarrow l=\frac{k}{n}\)
The total length of the string is
⇒ l = l1 + l2 + l3
Substitute the value of l, l1, l2, and l3 in the above equation, we get
\(\Rightarrow \frac{k}{n} = \frac{k}{n_1}+\frac{k}{n_2}+\frac{k}{n_3}\)
\(\Rightarrow \frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)
Wave Speed on a Stretched String Question 3:
The wave velocity in a guitar string of length l and radius r, made of a material of density ρ, is v. If another string of same length and radius, but made of a material of density 3ρ, is used, the wave velocity is nv, under the same tension. The value of n is
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 3 Detailed Solution
CONCEPT:
The wave velocity of a string under Tension T and mass per unit length μ
⇒ v = \(\sqrt{\frac{T}{μ} } \)
EXPLANATION:
For the first string,
Length = l, radius = r, density = ρ
The volume of the wire = πr2L
∴ density = mass/volume
⇒ volume = mass/density = m1/ρ
So, we have, πr2L = m1/ρ
⇒ m1/L = πr2ρ = mass per unit length = μ
So, velocity v = \(\sqrt{\frac{T}{\pi r^2\rho} } \) ---- (1)
Similarly for the second material of mass m2
⇒ m2/L = πr2(3ρ)
So, velocity v' = nv = \(\sqrt{\frac{T}{\pi r^ 2 3\rho} } = \frac{1}{\sqrt{3}} v = 0.58 v\)
∴ n = 0.58
Hence the correct answer is option 1.
Wave Speed on a Stretched String Question 4:
A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 × 104 N. When the tension is changed to T, the velocity changed to v/2. The value of T is close to :
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 4 Detailed Solution
CONCEPT:
For a transverse wave the velocity of the transverse wave
\(⇒v = \sqrt{\frac{T}{μ}} \)
Where, T = tension and μ = mass per unit length
∴ T = μ v2
EXPLANATION:
We know the relation between T and v
⇒ T ∝ v2
(Where μ is a constant)
⇒ T1/T2 = (v1/v2)2
∴ T2 = T1/(v1/v2)2
Given,
v1 = v and v2 = v/2
⇒v1/v2 = 2
⇒T1 = 2.06 × 104 N
∴ T2 = T1/(v1/v2)2
= (2.06 × 104)/(2)2 = (2.06/4) × 104 = 5.15 × 103 N
Hence the correct answer is option 2.
Wave Speed on a Stretched String Question 5:
A transverse wave described by y = (0.02 m) sin[x + 30t] propagates on a stretched string having a linear mass density of 1.2 × 10-4 kg/m. The tension in the string is
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 5 Detailed Solution
Concept:
The linear velocity of a wave traveling in a stretched string is given by the equation:
\(v=\sqrt{\frac{T}{m/L}}\)
Where T is the tension in the stretched string, m is the mass of the string and L is the length of the string.
The general equation of a wave is y = A sin(ωt + kx + ϕ)
Where y is the displacement of the wave at time t, A is the amplitude, ϕ is the phase and k is the angular wavenumber.
Calculation:
Given:
m/L = 1.2 × 10-4 kg/m
y = (0.02 m) sin[x + 30t]
Comparing the general wave equation
The general equation of a wave is y = A sin(ωt + kx + ϕ)
A = 0.02 m, ω = 30 rad/s, k = 1.
Linear velocity:
\(v=\frac{\omega}{k} =\frac {30}{1}=30\;m/s\)
\(v=\sqrt{\frac{T}{m/L}}\)
\(30=\sqrt{\frac{T}{1.2×10^{-4}}}\)
T = 30 × 30 × 1.2 × 10-4 = 0.108 N
Wave Speed on a Stretched String Question 6:
Consider a gas with molar mass M. If the sound at frequency f is introduced to a tube of this gas temperature T then an internal acoustic standing wave is set up with nodes separated by L. The adiabatic constant (γ = Cp/Cv) is
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 6 Detailed Solution
Concept:
- The frequency of oscillation is defined as the number of oscillations in one second.
- The distance between two nodes in terms of the wavelength λ the string is \(L=\frac{λ}{2}\)
- The relation between the speed of the wave v, frequency f and wavelength λ is given by, v=f λ
- The speed of the wave in string in terms of adiabatic constant, \(v=\sqrt{\frac{\gamma RT}{M}} \)
Explanation:
Let's consider a gas with a molar mass M, f is the frequency of a sound and T is the temperature.
The distance between two nodes is \(L=\frac{λ}{2}\)
So, λ =2L
The speed of the wave is, \(v=\sqrt{\frac{\gamma RT}{M}}\)
\(fλ=2fL=\sqrt{\frac{\gamma RT}{M}}\)
\(4f^2L^2=\frac{\gamma RT}{M}\)
\(\gamma=\frac{4f^2L^2M}{RT}\)
Additional Information
- Isothermal expansion: A process in which temperature remains constant. For such a process, PV = constant
- Isobaric process: A process in which pressure remains constant. For this process,
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Wave Speed on a Stretched String Question 7:
Two identical nylon strings are held stretched and it was noted that the fundamental frequency of the strings is in the ratio 1 : 2. Then the ratio of tension in both the string is:
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 7 Detailed Solution
The correct answer is option 4) i.e. 1 : 4
CONCEPT:
- The fundamental frequency of a stretched string is given by the equation:
\(\nu =\frac{1}{2L} \sqrt {\frac{T}{μ}}\)
Where L is the length of the vibrating part of the string, T is the tension and μ is the linear density of the string.
EXPLANATION:
Given that:
The ratio of frequencies is v1 : v2 = 1 : 2
Since the wires are identical, they will have the same L and μ.
The fundamental frequency, ν ∝ √T
Ratio, ν1 : ν2 = √T1 : √T2
⇒ν12 : ν22 = T1 : T2
12 : 22 = 1 : 4
Wave Speed on a Stretched String Question 8:
1A string of 7 metre length has a mass of 0.035 kg. If tension in the string is 60.5 N, then speed of a wave on the string is
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 8 Detailed Solution
Concept:
- Transverse wave: The wave generated such that the particles oscillate in the direction perpendicular to the propagation of the wave is called the transverse wave.
- The transverse wave can be observed when we pull a tight string a bit.
- The Speed of Such Transverse wave is given as:
\(v = \sqrt{\frac{T}{\mu}}\)
Where T is Tension in the tight String and μ is mass per unit length of the string.
Calculation:
Given length of string = 7m
Mass of string = 0.35 kg
Tension in string = 60.5 N
Mass per unit length
\(μ = \frac{0.35kg}{7m}\)
⇒ μ = 0.05
Speed of wave
\(v = \sqrt{\frac{T}{\mu}}\)
\(v = \sqrt{\frac{60.5}{0.05}}\)
Wave Speed on a Stretched String Question 9:
A string vibrates with a frequency of 200 Hz. When its length is doubled and tension is altered, it begins to vibrate with a frequency of 300 Hz. The ratio of the new tension to the original tension is
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 9 Detailed Solution
CONCEPT:
The frequency of oscillation of a string stretched under tension T is given by
\(f =\frac{1}{2l}\sqrt{\frac{T}{μ}}{}\)
Where f = frequency of oscillation, n = number of loops, l = Length of string, T =Tension, and μ= linear density
CALCULATION:
Given - f1 = 200 Hz, f2 = 300 Hz, L1 = l and L2 = 2l
- The frequency of oscillation is given by
\(\Rightarrow f =\frac{1}{2l}\sqrt{\frac{T}{μ}}\)
Where T = Tension and μ = mass
From the above equation it is clear that
\(\Rightarrow f \propto \frac{\sqrt{T}}{L}\)
For the two vibrating strings, the above equation is given by
\(\Rightarrow f _{1}\propto \frac{\sqrt{T_{1}}}{L_{1}}\) ------ (1)
\(\Rightarrow f_{2} \propto \frac{\sqrt{T_{2}}}{L_{2}}\) ------ (2)
On dividing equation 1 and 2, we get
\(\Rightarrow \frac{f_{2}}{f_{1}} =\sqrt{\frac{T_{2}}{T_{1}}} \frac{L_{2}}{L_{1}}\)
Squaring both sides of the above equation and it can be rewritten for \(\frac{T_{2}}{T_{1}}\)
\(\Rightarrow \frac{T_{2}}{T_{1}} =\frac{(f_{2}L_{2})^{2}}{(f_{1}L_{1})^{2}}\)
Substituting the given values of f1 and f2 and L2 = 2l
\(\Rightarrow \frac{T_{2}}{T_{1}} =\frac{(300\times 2L_{1})^{2}}{(200\times L_{1})^{2}}\)
\(\Rightarrow \frac{T_{2}}{T_{1}} =\frac{9}{1}\)
- Hence, option 1 is the answer
Wave Speed on a Stretched String Question 10:
A wire under tension vibrates with a fundamental frequency of 600 Hz. If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become ______
Answer (Detailed Solution Below)
Wave Speed on a Stretched String Question 10 Detailed Solution
CONCEPT:
- Fundamental frequency: It is the lowest frequency of a periodic waveform. It is also known as natural frequency.
The fundamental frequency of a wire:
\(f =\frac{1}{2l} . \sqrt{T\over μ}\)
where f is the fundamental frequency, l is the length of the wire, T is the tension in the wire, and μ is the mass per unit length.
\(f =\frac{1}{2l} . \sqrt{T\over μ}\)
\(f =\frac{1}{2l} . \sqrt{T\over π r^2 ρ}\)
\(f =\frac{1}{2lr} . \sqrt{T\over π ρ}\)
CALCULATION:
Given that length is doubled and the radius is halved and tension has been one-ninth.
T' = T/9; l' = 2l; r' = 2r; initial frequency f = 600 Hz
Fundamental frequency:
\(f =\frac{1}{2lr} . \sqrt{T\over π ρ}\)
in this formula π and ρ are constant.
\({f'\over f} =\frac{l'r'}{lr} . \sqrt{T'\over T}\)
\({f'\over f} =\frac{(l/2)(2r)}{lr} . \sqrt{T/9\over T}\)
\({f'\over f} ={1 \over 3}\)
f'/600 = 1/3
\(f' = 600/3 = 200 Hz\)
So the correct answer is option 1.