Ahorizontal 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.

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)

higher potential

explanation:

because it is in a high speed so it must move into a of high potential