Q: A transverse wave travelling on a stretched string is represented by the equation y=(2x−6.2t)2+202. Then,

vA= Coefficient of x Coefficient of t=26.2=3.1 m/s=202=0.1 m(a) velocity of the wave is 3.1 m/s(b) amplitude of the wave is 0.1 m

Q: For energy density, power and intensity of any wave choose the correct options.

(a) u= energy density =21ρω2A2(d) I=SP

Q: For the transverse wave equation y=Asin(πx+πt), choose the correct options at t=0

y=Asin(πx+πt)vP=∂t∂y=πAcos(πx+πt)aP=∂t2∂2y=−π2Asin(πx+πt)Now, substitute t=0 and given value of x.Since, ωt and kx are of same sign, hence the wave is travelling in negative x-direction.(a) points at x=0 and x=1 are at mean positions(b) points at x=0.5 and x=1.5 have maximum accelerations(c) points at x=0.5 and x=1.5 are at rest(d) the given wave is travelling in negative x-direction

Q: In the wave equation,y=Asina2π(x−bt)

Speed of wave= Coefficient of x Coefficient of t=1b=bλ=k2π=(2π/a)2π=a

Q: In the wave equation,y=Asin2π(ax−bt)

Speed of wave λT= Coefficient of x Coefficient of t=1/a1/b=ba=k2π=2π/a2π=a=ω2π=2π/b2π=b(a) speed of wave is a/b(c) wavelength of wave is a(d) time period of wave is b

Q: For the wave equation,y=(4 cm)sin[πt+2πx]Here, t is in second and x in metres.

(a)→q,r(b) → p,s(c) →q,r(d)→p,svP=∂t∂y=(4π)cm/scos[πt+2πx]aP=∂t2∂2y=(−4π2)cm/s2sin[πt+2πx]Now substitute the values of t and x.

Q: y-x graph of a transverse wave at a given instant is shown in figure. Match the following two columns.

(a) →s(b) →p(c) →S(d) →rFor velocity, vP=−v∂x∂y ∂x∂y= Slope of y−x graph.Sign of v is not given in the question.Hence, direction of vP cannot be determined.For particle acceleration,aP∝−yi.e. aP and y are away in opposite directions If y=0, then aP=0

Q: For a travelling wave, match the following two columns

(a)→s(b) → p,q(c) →s(d)→sEnergy density =21ρω2A2= Energy per unit volumePower = energy transfer per unitTime =21ρω2A2SvIntensity = energy transfer per unit per unit area=21ρω2A2v=SPWave number =λ1

Question 1

A transverse wave travelling on a stretched string is represented by the equation y=(2x−6.2t)2+202​. Then,

Accepted Answer

vA​= Coefficient of x Coefficient of t​=26.2​=3.1 m/s=202​=0.1 m​(a) velocity of the wave is 3.1 m/s(b) amplitude of the wave is 0.1 m

Question 2

For energy density, power and intensity of any wave choose the correct options.

Accepted Answer

(a) u= energy density =21​ρω2A2(d) I=SP​

Question 3

For the transverse wave equation y=Asin(πx+πt), choose the correct options at t=0

Accepted Answer

y=Asin(πx+πt)vP​=∂t∂y​=πAcos(πx+πt)aP​=∂t2∂2y​=−π2Asin(πx+πt)Now, substitute t=0 and given value of x.Since, ωt and kx are of same sign, hence the wave is travelling in negative x-direction.(a) points at x=0 and x=1 are at mean positions(b) points at x=0.5 and x=1.5 have maximum accelerations(c) points at x=0.5 and x=1.5 are at rest(d) the given wave is travelling in negative x-direction

Question 4

In the wave equation,y=Asina2π​(x−bt)

Accepted Answer

Speed of wave= Coefficient of x Coefficient of t​=1b​=bλ=k2π​=(2π/a)2π​=a

Question 5

In the wave equation,y=Asin2π(ax​−bt​)

Accepted Answer

Speed of wave λT​= Coefficient of x Coefficient of t​=1/a1/b​=ba​=k2π​=2π/a2π​=a=ω2π​=2π/b2π​=b​(a) speed of wave is a/b(c) wavelength of wave is a(d) time period of wave is b

Question 6

Corresponding to y t- graph of a transverse harmonic wave shown in figurechoose the correct options at same position.

Accepted Answer

y−t graph is sine graph. Therefore, v−t graph is cos graph and a−t is − sine graph asvP​=∂t∂y​ and aP​=∂t2∂2y​

Question 7

For the wave equation,y=asin(bt−cx)Match the following two columns.

Accepted Answer

(a) →r(b) →s(c) →p(d) →s(a) Wave speed = Coefficient of x Coefficient of t​=cb​(b) Maximum particle speed =ωA=(b)(a)(c) f=2πω​=2πb​(d) λ=k2π​=c2π​

Question 8

For the wave equation,y=(4 cm)sin[πt+2πx]Here, t is in second and x in metres.

Accepted Answer

(a)→q,r(b) → p,s(c) →q,r(d)→p,svP​=∂t∂y​=(4π)cm/scos[πt+2πx]aP​=∂t2∂2y​=(−4π2)cm/s2sin[πt+2πx]Now substitute the values of t and x.

Question 9

y-x graph of a transverse wave at a given instant is shown in figure. Match the following two columns.

Accepted Answer

(a) →s(b) →p(c) →S(d) →rFor velocity, vP​=−v∂x∂y​ ∂x∂y​= Slope of y−x graph.Sign of v is not given in the question.Hence, direction of vP​ cannot be determined.For particle acceleration,aP​∝−yi.e. aP​ and y are away in opposite directions If y=0, then aP​=0

Question 10

For a travelling wave, match the following two columns

Accepted Answer

(a)→s(b) → p,q(c) →s(d)→sEnergy density =21​ρω2A2= Energy per unit volumePower = energy transfer per unitTime =21​ρω2A2SvIntensity = energy transfer per unit per unit area=21​ρω2A2v=SP​Wave number =λ1​

Question 11

Match the following two columns

Accepted Answer

(a) → p,r(b) → p(c) → q,s(d) → p,svP​=∂t∂y​,aP​=∂t2∂2y​If ωt and kx are of same sign, then wave travel in negative x-direction.If they are of opposite signs, then wave travels in positive direction.

Question 12

The figure shows a snap photograph of a vibrating string at t=0. The particle P is observed moving up with velocity 203​ cm/s. The tangent at P makes an angle 60∘ with x-axis.y(a) Find the direction in which the wave is moving.(b) Write the equation of the wave.(c) The total energy carried by the wave per cycle of the string. Assuming that the mass per unitlength of the string is 50 g/m.

Accepted Answer

(a) vP​=−v(dxdy​)As vP​ and (slope) )P​ are both positive, v must be negative. Hence, the wave is moving in negative x-axis.(b) y=Asin(ωt−kx+ϕ)k=λ2π​=2π​ cm−1A=4×10−3 m=0.4 cmAt t=0,x=0, slope dxdy​=+ ve ∴vP​=−v( slope )=+ veFurther at t=0,x=0,y=+ ve ∴ϕ=4π​Further, 203​=−vtan60∘v=−20 cm/s∴f=λv​=5 Hz∴ω=2πf=10π∴y=(0.4 cm)sin(10πt+2π​x+4π​)(c) P=2π2A2f2μv∴ Energy carried per cycleE=PT=fP​=2π2A2fμvSubstituting the values, we haveE=1.6×10−5 J

Question 13

A long string having a cross-sectional area 0.80 mm2 and density 12.5 g/cm3 is subjected to a tension of 64 N along the positive x-axis. One end of this string is attached to a vibrator at x=0 moving in transverse direction at a frequency of 20 Hz. At t=0, the source is at a maximum displacement y=1.0 cm.a. Find the speed of the wave travelling on the string.b.Write the equation for the wave.c.What is the displacement of the particle of the string at x=50 cm at time t=0.05 s ?d.What is the velocity of this particle at this instant?

Accepted Answer

(a) v​=ρST​​=12.5×103×0.8×10−664​​=80 m/s​(b) ω=2πf=2π(20)=40πrad/s​k=vω​=8040π​=2π​ m−1∴y=(1.0 cm)cos[(40πs−1)t−(2π​m−1)x]​(c) Substituting x=0.5 m and t=0.05 s, we gety=2​1​ cm(d) Particle velocity at time t.vP​​=∂t∂y​=−(40πcm/s)sin[(40πs−1)t−(2π​ m−1)x]​Substituting x=0.5 m and t=0.05 s, we getvP​=89 cm/s

Question 14

One end of each of two identical springs, each of force constant 0.5 N/m are attached on the opposite sides of a wooden block of mass 0.01 kg. The other ends of the springs are connected to separate rigid supports such that the springs are unstretched and are collinear in a horizontal plane. To the wooden piece is fixed a pointer which touches a vertically moving plane paper. The wooden piece kept on a smooth horizontal table is now displaced by 0.02 m along the line of springs and released. If the speed of paper is 0.1 m/s, find the equation of the path traced by the pointer on the paper and the distance between two consecutive maximas on this path.

Accepted Answer

keff​=2kfv​=1.0 N/m=2π1​mkeff​​​=2π10​s−1=0.1 m/s,A=0.02 m​λ=fv​=(2π10​)0.1​=1002π​ my=Acos(λ2π​)(vt−x)=0.02cos100(0.1t−x)=0.02cos(10t−100x)mThe distance between two successive maxima=λ=1002π​=0.0628 m

Question 15

A wave pulse is travelling on a string with a speed v towards the positive x-axis. The shape of the string at t=0 is given by y(x)=Asin(x/a), where A and a are constants.(a) What are the dimensions of A and a ?(b) Write the equation of the wave for a general time t, if the wave speed is v.

Accepted Answer

(a) Dimensions of A and Y are same.Similarly, dimensions of a and x are same,  L,L(b) As the wave is travelling towards positivex-axis, there should be negative sign between term of x and term of t.Further, speed of wavev= Coefficient of x Coefficient of t​∴ Coefficient of t=(v)× coefficient of xy(x,t)=Asin(ax−vt​)

Question 16

Figure shows a plot of the transverse displacement of the particle of a string at t=0 through which a travelling wave is passing in the positive x-direction. 'The wave speed is 20 cm/s. Find (a) the amplitude (b) the wavelength (c) the wave number and (d) the frequency of the wave.

Accepted Answer

From the given figure, we can see that(a) Amplitude, A=1.0 mm(b) Wavelength, λ=4 cm(c) Wave number, k=λ2π​=1.57 cm−1≈1.6 cm−1(d) Frequency, f=λv​=420​=5 Hz

Question 17

Two wires of different densities but same area of cross-section are soldered together at one end and are stretched to a tension T. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.

Accepted Answer

v=ρST​​ or v∝ρ​1​∴v2​v1​​=ρ1​ρ2​​​∴ρ2​ρ1​​=(v1​v2​​)2=(21​)2=41​

Question 18

Two long strings A and B, each having linear mass density 1.2×10−2 kg/m are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x=0. Wave pulses are produced on the strings at the left ends at t=0 on string A and at t=20 ms on string B. When and where will the pulse on B overtake that on A ?

Accepted Answer

v1​v2​​=μ1​T1​​​=1.2×10−24.8​​=20 m/s=μ2​T2​​​=1.2×10−27.5​​=25 m/s​Pulses will meet when xA​=xB​or20tt​=25(t−0.02)=0.1 s​∴t=0.1 sand xA​ or xB​=20×0.1=2 m

Question 19

A sinusoidal transverse wave travels on a string. The string has length 8.00 m and mass 6.00 g. The wave speed is 30.0 m/s and the wavelength is 0.200 m. (a) If the wave is to have an average power of 50.0 W, what must be the amplitude of the wave? (b) For this same string, if the amplitude and wavelength are the same as in part (a) what is the average power for the wave if the tension is increased such that the wave speed is doubled?

Accepted Answer

(a)P=21​ρω2A2Sv or A=ω1​ρSv2P​​​orA=ω1​ρSv2P​​Here, ρS=μ= mass per unit lengthP=21​ρω2A2Sv or A=ω1​ρSv2P​​​Substituting these values in Eq. (i), we haveA​=2π×300.2​6×10−3×302×50×8​​=0.0707 m=7.07 cm​(b) P∝vω2P∝v(v2)P∝v3​When wave speed is doubled, then power will become eight times.

Question 20

A uniform rope with length L and mass m is held at one end and whirled in a horizontal circle with angular velocity ω. You can ignore the force of gravity on the rope. Find the time required for a transverse wave to travel from one end of the rope to the other.
Hint : ∫a2−x2​dx​=sin−1(ax​)

Accepted Answer

−dT=(dm)xω2=(Lm​dx)xω2 or −∫0T​dT=Lm​ω2∫x=Lx​xdx∴−T=Lm​ω2(2x2​−2L2​) or T=2Lmω2​(L2−x2)v=μT​​=m/L2Lmω2​(L2−x2)​​​ or dtdx​∴∫0t​dt∴t​=ω2L2−x2​​=2​ω​L2−x2​=ω2​​∫0L​L2−x2​dx​=ω2​​[sin−1(Lx​)]0L​=ω2​​⋅2π​=2​ωπ​​

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Solutions for Waves And Thermodynamics by DC Pandey

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