Wave Speed on a Stretched String MCQ Quiz - Objective Question with Answer for Wave Speed on a Stretched String - Download Free PDF

Concept:

Length of the String Under Tension:

• The extension of a string under tension is directly proportional to the applied tension and the original length.
• The formula for the length of the string under tension is: L = L₀ (1 + ΔL/L₀) = L₀ + (T / k), where T is the tension and k is the constant of proportionality.
• From the two given tensions, we can use the proportionality to form equations for the lengths.

Calculation:

Let the original length of the string be L₀.

When the tension is 6 N, the length is P: P = L₀ + (6 / k)

When the tension is 8 N, the length is Q: Q = L₀ + (8 / k)

Subtract the two equations: Q - P = (L₀ + 8 / k) - (L₀ + 6 / k)

Q - P = (2 / k)

k = 2 / (Q - P)

Substitute this value of k into one of the equations, say P = L₀ + (6 / k):

P = L₀ + (6 / (2 / (Q - P))) = L₀ + 3(Q - P)

L₀ = P - 3(Q - P) = 4P - 3Q

∴ The original length of the string is 4P - 3Q. Option 4) is correct.

Concept:

Wave Speed in a Stretched String:

• The speed of a transverse wave in a stretched string is given by the formula:
  v = √(T / μ)
  • T = tension in the string,
    μ = linear mass density of the string.
• For two strings with the same length and radius, but different densities, the speed of the wave is inversely proportional to the square root of the density of the material of the string.
• If the density of string B is 2% more than that of string A, the speed of the wave in string B will be modified accordingly.

Calculation:

Given,

Speed of wave in string A, v = √(T / μ_A)

Density of string B, ρ_B = 1.02 × ρ_A (since 2% more than density of A)

Speed of wave in string B, v_B = √(T / μ_B)

Since the tension is the same for both strings and their radii are the same, the linear mass density is directly related to the material density. Therefore, μ_B = 1.02 × μ_A.

Thus, the speed of the wave in string B will be:

v_B = √(T / (1.02 × μ_A))

v_B = v / √(1.02)

∴ The speed of the wave in string B is v / √(1.02).
Hence, the correct option is 4) v / √(1.02).

Concept:

Wave Equation and Tension in a String:

• The wave equation for a string is given by: y = A sin( 2π ( t / T - x / λ ) )
• Where:
  • A = Amplitude of the wave
  • T = Period of the wave
  • λ = Wavelength of the wave
  • t = Time
  • x = Position along the string
• For a string under tension, the tension (T) is related to the wave speed and linear mass density (μ) by the formula: T = μ v² = μ ( ω / k )²
• Where:
  • μ = Linear mass density (kg/m)
  • v = Wave speed (m/s)
  • ω = Angular frequency (rad/s)
  • k = Wave number (rad/m)

Calculation:

Given,

Amplitude, A = 0.02 m

Linear mass density, μ = 0.04 kg/m

The wave number (k) is given by: k = 2π / λ = 2π / 0.50 = 2π / 0.50 rad/m

The angular frequency (ω) is given by: ω = 2π / T = 2π / 0.004 rad/s

The tension in the string is:

T = μ ( ω / k )² = 0.04 × ( (2π / 0.004) / (2π / 0.50) )² = 6.25 N

∴ The tension in the string is 6.25 N.

Concept Used:

The minimum time interval (Δt) that allows for constructive interference occurs when a wave pulse travels from one end of the string to the other and back. This time is the round-trip time, which can be determined by calculating the wave speed on the string and using it to find the time for the wave to travel the round-trip distance (2L).

The wave speed v on the string is given by the formula:

v = √(T / μ)

Where:

• T is the tension in the string, given as 2.5 N.
• μ is the linear mass density of the string.

The linear mass density μ can be calculated as:

μ = m / L

Where:

m is the mass of the string, given as 10^-3 kg.

L is the length of the string, given as 25 cm (which we convert to meters, so 0.25 m).

Calculation:

Given:

Mass of the string, m = 10^-3 kg

Length of the string, L = 0.25 m

Tension in the string, T = 2.5 N

First, we calculate the linear mass density μ:

⇒ μ = (10^-3 kg) / (0.25 m) = 4 × 10^-3 kg/m

Now, substitute the values of T and μ into the wave speed formula:

⇒ v = √(2.5 N / 4 × 10^-3 kg/m) = √(625 m²/s²) = 25 m/s

Next, we calculate the time for the wave pulse to travel the round-trip distance (2L):

⇒ Δt = (2L) / v

⇒ Δt = (2 × 0.25 m) / (25 m/s) = 0.5 m / 25 m/s = 0.02 s = 20 ms

∴ Option 4: 20 ms

Calculation:

\(v_{\text{max}} = a \omega = \frac{v}{10} = \frac{10}{10} = 1 \, \text{m/sec}\)

\(\Rightarrow a \omega = a \times 2\pi n = 1 \Rightarrow n = \frac{10^{3}}{2\pi}\)

\( \Rightarrow n = \frac{10^{3}}{2\pi}\)        (\(∴ a = 10^{-3} \, \text{m}\))

Since, \(v = n \lambda \)

\(\Rightarrow \lambda = \frac{v}{n} = \frac{10}{10^{3}/2\pi} = 2\pi \times 10^{-2} \, \text{m}\)

∴ Correct options are 1 & 3.

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

\(\Rightarrow \frac{k}{n} = \frac{k}{n_1}+\frac{k}{n_2}+\frac{k}{n_3}\)

\(\Rightarrow \frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)

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,

Tension in String T = 60N

Mass of String m = 5.0 × 10–3 kg

Length of String l = 0.72 m

Mass Per Unit Length of String \(\mu = \frac{5\times 10^{-3}kg}{0.72m}\)

⇒μ = 6.67 × 10^-3 kg 

Speed \(v = \sqrt{\frac{T}{\mu }}\)

⇒ \(v = \sqrt{\frac{60}{6.67\times10 ^{-3} }} m/s \)

Solving it, we will get the approx value of v = 93 m/s. 

So, Option 3 is the correct answer.

CONCEPT:

• Simple Harmonic Motion (SHM): Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
  • Example: Motion of an undamped pendulum, undamped spring-mass system.

The speed of a transverse waves on a stretched string is given by:

\({\rm{v}} = \sqrt {\frac{{\rm{T}}}{{\rm{\mu }}}}\)

Where v is the velocity of the wave, T is the tension in the string; μ is mass per unit length.

EXPLANATION:

The speed of transverse waves on a stretched string is given by v = √(T/X).

1. Here X is mass per unit length or linear density of string. So option 1 is correct.
2. Bulk modulus of elasticity (B): It is the ratio of Hydraulic (compressive) stress (p) to the volumetric strain (ΔV/V).
3. Young’s modulus: Young's modulus a modulus of elasticity, applicable to the stretching of wire, etc., equal to the ratio of the applied load per unit area of the cross-section to the increase in length per unit length.
4. Density: The mass per unit volume is called denisty.

CONCEPT:

• Simple Harmonic Motion (SHM): Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
• Example: Motion of an undamped pendulum, undamped spring-mass system.

\({\rm{v}} = \sqrt {\frac{{\rm{T}}}{{\rm{\mu }}}}\) , where v is the velocity of the wave, T is the tension in the string; μ is mass per unit length.

CALCULATION:

Given length l = 2 m, mass of 2 m string m = 10 g, wave velocity v = 40 m/s.

\({\rm{\mu }} = \frac{{\rm{m}}}{{\rm{l}}} = \frac{{10}}{2} = \)  5 g/m = 5 × 10^-3 kg/m. from formula,

\({\rm{v}} = \sqrt {\frac{{\rm{T}}}{{\rm{\mu }}}} \) ;

\(40 = \sqrt {\frac{T}{{5 \times {{10}^{ - 3}}}}} \) ⇒ 40² × 5 × 10^-3 = 8 N.

T = 8 N.

Concept:

• Simple Harmonic Motion (SHM): Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
  • Example: Motion of an undamped pendulum, undamped spring-mass system.

The speed of a transverse wave on a stretched string is given by:

\({\rm{v}} = \sqrt {\frac{{\rm{T}}}{{\rm{μ }}}}\,\)

Where v is the velocity of the wave, T is the tension in the string; μ is mass per unit length.

Calculation:

Let μ is the mass per unit length of the roap.

The speed of transverse waves on a stretched string is given by v = √(T/μ).

At a distance x from the lower end, we will find tension which will be

T = μ g x

So, using

\({\rm{v}} = \sqrt {\frac{{\rm{T}}}{{\rm{μ }}}}\,\)  (Speed of transverse wave on a string)

\({\rm{v}} = \sqrt {\frac{{\rm{\mu \,g\,x}}}{{\rm{μ }}}}\,\)

v =\(\sqrt {gx}\,\)

EXPLANATION:

→Speed of the transverse wave in any string \(v = \sqrt{\frac{T}{μ}} \) 

Where, T = tension in the string, μ = mass per unit length

Given: 

mass, m = 2.5 kg, length, l = 20.0 m

∴ μ = m/l = 2.5/20 = 0.125 kg/m

Hence speed, \(v = \sqrt{\frac{200}{0.125}}\) 

v = √1600 = 40 m/s

∴ Time taken for the wave to travel from one end to the other end

= Time taken for the wave to travel 20 m

t = distance/velocity = 20/40 = 0.5 s

So, the correct answer is option (2).

CONCEPT:

• Fundamental frequency: It is the lowest frequency of a periodic waveform. It is also known as natural frequency.

The fundamental frequency of a sonometer 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.

EXPLANATION:

The frequency is given by:

\(f =\frac{1}{2l} . \sqrt{T\over μ}\)

Given that:

Tension is made four times,

So new tension (T') = 4T

The new frequency is given by:

\(f' =\frac{1}{2l} . \sqrt{4T\over μ} =2 \times \frac{1}{2l} . \sqrt{T\over μ} = 2f\)

• Thus the frequency becomes 2 times. So option 1 is correct.

Explanation:

Given:

String A and B are of the same material and produce a frequency which is 6 Hz.

Therefore, The difference between f_A and f_B is 6 Hz

i.e, fA – fB = 6 Hz ---(1)

Where f_A is the frequency of A and f_B is the frequency of B respectively.

If tension decreases, f_B decreases and becomes f_B.

Now, the difference between f_A and f_B = 7 Hz (increases), which means that the frequency of string A is greater than the frequency of string B.

So, f_A > f_B

As we know, f_A = 530 Hz

Putting the value of f_A in equation (1) we get;

530 – f_B = 6 

⇒ -f_B = 6-530

⇒ -f_B = -524

⇒ f_B = 524 Hz.

Hence, option 4) is the correct answer.

The correct answer is option 1) i.e. 7.

CONCEPT:

• Frequency in a stretched string:
  • A stretched string is always subjected to a tension force along the length of the string.
  • When a stretched string is vibrated, it produces transverse waves.

The frequency of the transverse wave (ν) is related to the tension in the string as follows:

 \(ν = \frac{1}{2L} \sqrt{\frac{T}{μ}}\)

Where L is the length of the string, T is the tension in the string and μ is the mass per unit length of the given string.

• Beats: Beats are the periodic fluctuations heard in the intensity of a sound when two sound waves of nearly identical frequencies interfere with one another.
  • ​The number of beats is found from the difference in frequencies of two interfering waves. 

​CALCULATION:

Frequency, \(ν = \frac{1}{2L} \sqrt{\frac{T}{μ}}\)

\(ν_1 = \frac {1}{2 \times 0.516} \sqrt{\frac{20}{10^{-3}}} = \) 137 Hz

\(ν_2 = \frac {1}{2 \times 0.491} \sqrt{\frac{20}{10^{-3}}} = \) 144 Hz

Beats = difference in frequencies = ν₂ - ν₁ = 144 - 137 = 7

CONCEPT:

• The stationary wave is also known as the standing wave.
  • It is a combination of two waves, with the same amplitude and the same frequency, moving in the opposite direction.
    • It is a result of interference.
• Harmonics of an instrument: A musical instrument has a set of natural frequencies at which it vibrates when a disturbance is introduced into it.
  • This set of natural frequencies are known as the harmonics of the instrument.

The Frequency of nth harmonic in the standing wave is given by:

\(ν_n=\frac{nv}{2L}\)

where n is nth harmonic, v is the speed of sound, L is the length string.

CALCULATION:

For second harmonic n = 2

Frequency of 2^nd harmonic in the standing wave

\(ν=\frac{nv}{2L}\)

\(ν=\frac{2v}{2L}\)

\(ν=\frac{v}{L}\)

So the correct answer is option 1.