BPHCT-137 Solved Assignment 2024

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BPHCT-137 Solved Assignment 2024 Sample Solution
bphct-137-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

bphct-137-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

BPHCT-137 Solved Assignment 2024 SS
PART A
  1. a) Obtain an expression for energy transported by a progressive wave on a taut string. Also show that the power transported by the wave is proportional to the wave velocity.
Answer:
The energy transported by a traveling wave on a taut string can be analyzed by considering the kinetic and potential energies of a small element of the string. Let’s derive the expressions for these energies and then calculate the total energy and power transported by the wave.

Kinetic Energy:

For a small element Δ x Δ x Delta x\Delta xΔx of the string, the kinetic energy ( d ( K . E . ) d ( K . E . ) d(K.E.)d(K.E.)d(K.E.)) is given by:
d ( K . E . ) = 1 2 m v 2 = 1 2 ( ρ Δ x ) ( ∂ y ∂ t ) 2 d ( K . E . ) = 1 2 m v 2 = 1 2 ( ρ Δ x ) ∂ y ∂ t 2 d(K.E.)=(1)/(2)mv^(2)=(1)/(2)(rho Delta x)((del y)/(del t))^(2)d(K.E.) = \frac{1}{2} m v^2 = \frac{1}{2} (\rho \Delta x) \left(\frac{\partial y}{\partial t}\right)^2d(K.E.)=12mv2=12(ρΔx)(∂y∂t)2
where ρ ρ rho\rhoρ is the mass per unit length of the string, and ∂ y ∂ t ∂ y ∂ t (del y)/(del t)\frac{\partial y}{\partial t}∂y∂t is the velocity of the element.
For a sinusoidal wave y ( x , t ) = a sin ⁡ ( ω t − k x ) y ( x , t ) = a sin ⁡ ( ω t − k x ) y(x,t)=a sin(omega t-kx)y(x, t) = a \sin(\omega t – kx)y(x,t)=asin⁡(ωt−kx), the velocity is:
∂ y ∂ t = a ω cos ⁡ ( ω t − k x ) ∂ y ∂ t = a ω cos ⁡ ( ω t − k x ) (del y)/(del t)=a omega cos(omega t-kx)\frac{\partial y}{\partial t} = a \omega \cos(\omega t – kx)∂y∂t=aωcos⁡(ωt−kx)
Substituting this into the kinetic energy expression:
d ( K . E . ) = 1 2 ( ρ Δ x ) [ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) ] d ( K . E . ) = 1 2 ( ρ Δ x ) a 2 ω 2 cos 2 ⁡ ( ω t − k x ) d(K.E.)=(1)/(2)(rho Delta x)[a^(2)omega^(2)cos^(2)(omega t-kx)]d(K.E.) = \frac{1}{2} (\rho \Delta x) \left[a^2 \omega^2 \cos^2(\omega t – kx)\right]d(K.E.)=12(ρΔx)[a2ω2cos2⁡(ωt−kx)]

Potential Energy:

The potential energy ( d ( P . E . ) d ( P . E . ) d(P.E.)d(P.E.)d(P.E.)) arises due to the stretching of the string and is given by:
d ( P . E . ) = 1 2 T ( ∂ y ∂ x ) 2 Δ x d ( P . E . ) = 1 2 T ∂ y ∂ x 2 Δ x d(P.E.)=(1)/(2)T((del y)/(del x))^(2)Delta xd(P.E.) = \frac{1}{2} T \left(\frac{\partial y}{\partial x}\right)^2 \Delta xd(P.E.)=12T(∂y∂x)2Δx
where T T TTT is the tension in the string.
For the sinusoidal wave, the derivative with respect to x x xxx is:
∂ y ∂ x = − a k cos ⁡ ( ω t − k x ) ∂ y ∂ x = − a k cos ⁡ ( ω t − k x ) (del y)/(del x)=-ak cos(omega t-kx)\frac{\partial y}{\partial x} = -a k \cos(\omega t – kx)∂y∂x=−akcos⁡(ωt−kx)
Substituting this into the potential energy expression:
d ( P . E . ) = 1 2 ( T Δ x ) [ a 2 k 2 cos 2 ⁡ ( ω t − k x ) ] d ( P . E . ) = 1 2 ( T Δ x ) a 2 k 2 cos 2 ⁡ ( ω t − k x ) d(P.E.)=(1)/(2)(T Delta x)[a^(2)k^(2)cos^(2)(omega t-kx)]d(P.E.) = \frac{1}{2} (T \Delta x) \left[a^2 k^2 \cos^2(\omega t – kx)\right]d(P.E.)=12(TΔx)[a2k2cos2⁡(ωt−kx)]

Total Energy:

The total mechanical energy ( d E d E dEdEdE) of the element Δ x Δ x Delta x\Delta xΔx is the sum of the kinetic and potential energies:
d E = d ( K . E . ) + d ( P . E . ) = 1 2 Δ x a 2 cos 2 ⁡ ( ω t − k x ) [ ρ ω 2 + T k 2 ] = 1 2 Δ x a 2 cos 2 ⁡ ( ω t − k x ) [ 2 ρ ω 2 ] (since v = ω k and T = ρ v 2 ) = ρ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) Δ x d E = d ( K . E . ) + d ( P . E . ) = 1 2 Δ x a 2 cos 2 ⁡ ( ω t − k x ) ρ ω 2 + T k 2 = 1 2 Δ x a 2 cos 2 ⁡ ( ω t − k x ) 2 ρ ω 2 (since v = ω k and T = ρ v 2 ) = ρ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) Δ x {:[dE=d(K.E.)+d(P.E.)],[=(1)/(2)Delta xa^(2)cos^(2)(omega t-kx)[rhoomega^(2)+Tk^(2)]],[=(1)/(2)Delta xa^(2)cos^(2)(omega t-kx)[2rhoomega^(2)]quad(since v=(omega )/(k)” and “T=rhov^(2))],[=rhoa^(2)omega^(2)cos^(2)(omega t-kx)Delta x]:}\begin{aligned} dE &= d(K.E.) + d(P.E.) \\ &= \frac{1}{2} \Delta x a^2 \cos^2(\omega t – kx) \left[\rho \omega^2 + T k^2\right] \\ &= \frac{1}{2} \Delta x a^2 \cos^2(\omega t – kx) \left[2 \rho \omega^2\right] \quad \text{(since } v = \frac{\omega}{k} \text{ and } T = \rho v^2) \\ &= \rho a^2 \omega^2 \cos^2(\omega t – kx) \Delta x \end{aligned}dE=d(K.E.)+d(P.E.)=12Δxa2cos2⁡(ωt−kx)[ρω2+Tk2]=12Δxa2cos2⁡(ωt−kx)[2ρω2](since v=ωk and T=ρv2)=ρa2ω2cos2⁡(ωt−kx)Δx
The linear energy density ( D ( x , t ) D ( x , t ) D(x,t)D(x, t)D(x,t)) of the wave is then:
D ( x , t ) = d E Δ x = ρ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) D ( x , t ) = d E Δ x = ρ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) D(x,t)=(dE)/(Delta x)=rhoa^(2)omega^(2)cos^(2)(omega t-kx)D(x, t) = \frac{dE}{\Delta x} = \rho a^2 \omega^2 \cos^2(\omega t – kx)D(x,t)=dEΔx=ρa2ω2cos2⁡(ωt−kx)

Energy Contained in One Wavelength:

The total mechanical energy ( E E EEE) contained in one wavelength ( λ λ lambda\lambdaλ) of the wave is obtained by integrating D ( x , t ) D ( x , t ) D(x,t)D(x, t)D(x,t) over one wavelength:
E = ∫ 0 λ D ( x , t ) d x = ρ a 2 ω 2 ∫ 0 λ cos 2 ⁡ ( k x ) d x = 1 2 ρ a 2 ω 2 λ E = ∫ 0 λ   D ( x , t ) d x = ρ a 2 ω 2 ∫ 0 λ   cos 2 ⁡ ( k x ) d x = 1 2 ρ a 2 ω 2 λ E=int_(0)^(lambda)D(x,t)dx=rhoa^(2)omega^(2)int_(0)^(lambda)cos^(2)(kx)dx=(1)/(2)rhoa^(2)omega^(2)lambdaE = \int_0^\lambda D(x, t) dx = \rho a^2 \omega^2 \int_0^\lambda \cos^2(kx) dx = \frac{1}{2} \rho a^2 \omega^2 \lambdaE=∫0λD(x,t)dx=ρa2ω2∫0λcos2⁡(kx)dx=12ρa2ω2λ

Power Transported by the Wave:

The power ( P P PPP) of the wave is the energy averaged over one time period ( T T TTT) and is given by:
P = E T = E ( λ / v ) = 1 2 ρ a 2 ω 2 v = 2 π 2 ρ a 2 f 2 v P = E T = E ( λ / v ) = 1 2 ρ a 2 ω 2 v = 2 π 2 ρ a 2 f 2 v P=(E)/(T)=(E)/((lambda//v))=(1)/(2)rhoa^(2)omega^(2)v=2pi^(2)rhoa^(2)f^(2)vP = \frac{E}{T} = \frac{E}{(\lambda / v)} = \frac{1}{2} \rho a^2 \omega^2 v = 2 \pi^2 \rho a^2 f^2 vP=ET=E(λ/v)=12ρa2ω2v=2π2ρa2f2v
where ω = 2 π f ω = 2 π f omega=2pi f\omega = 2 \pi fω=2πf and f f fff is the frequency of the wave.

Conclusion:

The energy transported by a traveling wave on a taut string is distributed between kinetic and potential energies, with the total energy contained in one wavelength given by 1 2 ρ a 2 ω 2 λ 1 2 ρ a 2 ω 2 λ (1)/(2)rhoa^(2)omega^(2)lambda\frac{1}{2} \rho a^2 \omega^2 \lambda12ρa2ω2λ. The power transported by the wave is proportional to the wave velocity, as well as the square of its amplitude and frequency.
b) The fundamental frequency of a string instrument is 580 H z 580 H z 580Hz580 \mathrm{~Hz}580 Hz. (i) Calculate the frequency of the first and third harmonics generated on it. (ii) If the tension in the string is doubled, calculate the new fundamental frequency.
Answer:
To address this problem, we’ll break it down into two parts as requested:

Part (i): Frequency of the First and Third Harmonics

The fundamental frequency of a string instrument is given as f 1 = 580 H z f 1 = 580 H z f_(1)=580Hzf_1 = 580 \, \mathrm{Hz}f1=580Hz. The harmonics of a string instrument are integer multiples of the fundamental frequency. Thus, the frequency of the n n nnnth harmonic can be calculated using the formula:
f n = n ⋅ f 1 f n = n ⋅ f 1 f_(n)=n*f_(1)f_n = n \cdot f_1fn=n⋅f1
  • For the first harmonic ( n = 1 n = 1 n=1n=1n=1), this is simply the fundamental frequency itself, f 1 = 580 H z f 1 = 580 H z f_(1)=580Hzf_1 = 580 \, \mathrm{Hz}f1=580Hz.
  • For the third harmonic ( n = 3 n = 3 n=3n=3n=3), the frequency is f 3 = 3 ⋅ f 1 f 3 = 3 ⋅ f 1 f_(3)=3*f_(1)f_3 = 3 \cdot f_1f3=3⋅f1.
Let’s substitute the values to find f 3 f 3 f_(3)f_3f3:
f 3 = 3 ⋅ 580 H z f 3 = 3 ⋅ 580 H z f_(3)=3*580Hzf_3 = 3 \cdot 580 \, \mathrm{Hz}f3=3⋅580Hz
After Calculating, we get:
f 3 = 1740 H z f 3 = 1740 H z f_(3)=1740Hzf_3 = 1740 \, \mathrm{Hz}f3=1740Hz

Part (ii): New Fundamental Frequency if the Tension is Doubled

The fundamental frequency of a string can also be expressed in terms of the tension ( T T TTT) in the string, its length ( L L LLL), and its mass per unit length ( μ μ mu\muμ) as follows:
f 1 = 1 2 L T μ f 1 = 1 2 L T μ f_(1)=(1)/(2L)sqrt((T)/( mu))f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}f1=12LTμ
When the tension in the string is doubled ( T ′ = 2 T T ′ = 2 T T^(‘)=2TT’ = 2TT′=2T), the new fundamental frequency ( f 1 ′ f 1 ′ f_(1)^(‘)f_1′f1′) can be calculated using the modified tension:
f 1 ′ = 1 2 L T ′ μ = 1 2 L 2 T μ f 1 ′ = 1 2 L T ′ μ = 1 2 L 2 T μ f_(1)^(‘)=(1)/(2L)sqrt((T^(‘))/(mu))=(1)/(2L)sqrt((2T)/(mu))f_1′ = \frac{1}{2L} \sqrt{\frac{T’}{\mu}} = \frac{1}{2L} \sqrt{\frac{2T}{\mu}}f1′=12LT′μ=12L2Tμ
Since 2 T = 2 ⋅ T 2 T = 2 ⋅ T sqrt(2T)=sqrt2*sqrtT\sqrt{2T} = \sqrt{2} \cdot \sqrt{T}2T=2⋅T, we can relate the new fundamental frequency to the old one as follows:
f 1 ′ = 2 ⋅ f 1 f 1 ′ = 2 ⋅ f 1 f_(1)^(‘)=sqrt2*f_(1)f_1′ = \sqrt{2} \cdot f_1f1′=2⋅f1
Substituting the given fundamental frequency ( f 1 = 580 H z f 1 = 580 H z f_(1)=580Hzf_1 = 580 \, \mathrm{Hz}f1=580Hz) into the formula gives us:
f 1 ′ = 2 ⋅ 580 H z f 1 ′ = 2 ⋅ 580 H z f_(1)^(‘)=sqrt2*580Hzf_1′ = \sqrt{2} \cdot 580 \, \mathrm{Hz}f1′=2⋅580Hz
After Calculating, we get:
f 1 ′ ≈ 819.6 H z f 1 ′ ≈ 819.6 H z f_(1)^(‘)~~819.6Hzf_1′ \approx 819.6 \, \mathrm{Hz}f1′≈819.6Hz

Summary

  • The frequency of the first harmonic is 580 H z 580 H z 580Hz580 \, \mathrm{Hz}580Hz, which is the same as the fundamental frequency.
  • The frequency of the third harmonic is 1740 H z 1740 H z 1740Hz1740 \, \mathrm{Hz}1740Hz.
  • If the tension in the string is doubled, the new fundamental frequency becomes approximately 819.6 H z 819.6 H z 819.6Hz819.6 \, \mathrm{Hz}819.6Hz.
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