Physics

Question

A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed v, frequency v, amplitude A, and wavelength λ. Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) x= λ /2, (ii) x= λ /4 and (iii) x= λ /8 from the left-hand end of the string.

2 Answer

  • Answer:

    1)  both velocity and acceleration at x =  λ /2 are 0

    2)maximum velocity at x=λ /4 is -2Aω

      maximum acceleration at x=λ /4 is Aω²

    3)maximum velocity at x=λ /8 is √2Aω

    maximum acceleration at x=λ /8 is (-Aω²/√2)

    Explanation:

    When a string is tied at both ends and vibrated, we get standing waves.

    The general form of standing wave is,

    y(x,t) = 2A×sin(kx)×cos(ωt)   (displacement of string at x and at time t)

    where, A = amplitude of wave

    ω = 2πv , v = frequency of the wave

    k = 2π/λ ;  λ = wavelength of wave

    Speed V = ω/k.

    Now, differentiating y(x,t) by once with t gives velocity

    ⇒V(x,t) = -2Aω×sin(kx)×sin(ωt)

    And differentiating once again gives acceleration

    a(x,t) = -Aω²×sin(kx)×cos(ωt)

    1) x= λ/2

    ⇒V(x,t) = -2Aω×sin(k(λ /2))×sin(ωt)

                = -2Aω×sin(π)×sin(ωt)

                = 0

    a(x,t) = -Aω²×sin(kx)×cos(ωt)

    ⇒ a(x,t) = -Aω²×sin(k(λ /2))×cos(ωt)

                 = -Aω²×sin(π)×cos(ωt)

                 = 0

    2) x= λ /4

    ⇒V(x,t) = -2Aω×sin(kx)×sin(ωt)

                = -2Aω×sin(k(λ /4))×sin(ωt)

                = -2Aω×sin(π/2)×sin(ωt)

                 = -2Aω sin(ωt)

    ⇒ maximum velocity at x=λ /4 is -2Aω (minimum sin value is -1)

    a(x,t) = -Aω²×sin(kx)×cos(ωt)

            = -Aω²×cos(ωt)

    ⇒ maximum acceleration at x=λ /4 is Aω² (minimum cos value is -1)

    3) x= λ /8

    ⇒V(x,t) = -2Aω×sin(kx)×sin(ωt)

                 = -√2Aω×sin(ωt)

    ⇒ maximum velocity at x=λ /8 is √2Aω (minimum sin value is -1)

    a(x,t) = -Aω²×sin(kx)×cos(ωt)

             = (-Aω²/√2) × cos(ωt)

    ⇒ maximum acceleration at x=λ /8 is (-Aω²/√2) (minimum cos value is -1)

     

  • The velocity perpendicular to the radius vector is called the transverse velocity. The transverse acceleration is the acceleration produced by an inertial force acting across the body. The expression for the maximum transverse velocity and acceleration has been obtained for given values of x.

    Given information-

    A horizontal string tied at both ends is vibrating in its fundamental mode.

    Transverse velocity

    Velocity perpendicular to the radius vector is called the transverse velocity.

    The formula for the maximum transverse velocity can be given as,

    [tex]V_m=-2A\omega\sin(kx)\times \sin(\omega t)[/tex]

    Transverse acceleration

    Transverse acceleration is the acceleration produced by an inertial force acting across the body.

    [tex]a_m=-A\omega^2 \sin(kx)\times \cos(\omega t)[/tex]

    • 1) Maximum transverse velocity and maximum transverse acceleration when x is equal to the [tex]\lambda/2 [/tex] -

    [tex]V_m=-2A\omega\sin(k\dfrac{\lambda}{y} )\times \sin(\omega t)\\ V_m=-2A\omega\sin(\pi )\times \sin(\omega t)\\ V_m=0[/tex]

    [tex]a_m=-A\omega^2 \sin(k\dfrac{\lambda}{2} )\times \cos(\omega t)\\ a_m=-A\omega^2 \sin(\pi )\times \cos(\omega t)\\ a_m=0[/tex]

    • 2) Maximum transverse velocity and maximum transverse acceleration when x is equal to the [tex]\lambda/4 [/tex] -

    [tex]V_m=-2A\omega\sin(k\dfrac{\lambda}{4} )\times \sin(\omega t)\\ V_m=-2A\omega\sin(\dfrac{\pi}{2} )\times \sin(\omega t)\\ V_m=-2A\omega \times \sin(\omega t)\\ [/tex]

    [tex]a_m=-A\omega^2 \sin(k\dfrac{\lambda}{4} )\times \cos(\omega t)\\ a_m=-A\omega^2 \sin(\dfrac{\pi}{2} )\times \cos(\omega t)\\ a_m=-A\omega^2 \times \cos(\omega t)\\[/tex]

    • 3) Maximum transverse velocity and maximum transverse acceleration when x is equal to the [tex]\lambda/3 [/tex] -

    [tex]V_m=-2A\omega\sin(k\dfrac{\lambda}{8} )\times \sin(\omega t)\\ V_m=-{\sqrt{2} }{A\omega\times \sin(\omega t)} \\ [/tex]

    [tex]a_m=-A\omega^2 \sin(k\dfrac{\lambda}{4} )\times \cos(\omega t)\\ a_m=\dfrac{-A\omega^2\times \cos(\omega t)}{\sqrt{2} } \\ [/tex]

    Thus the maximum transverse velocity and maximum transverse acceleration at given points has been obtained.

    Learn more about transverse velocity and  transverse acceleration here;

    https://brainly.com/question/18394210

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