Question

A plucked violin string carries a traveling wave given by the equation f(x,t)=asin[b(x−ct)+ϕi], with a = 0.00580 m , b = 33.05 m−1 , and c = 245 m/s .
A) If f(0, 0) = 0, what is ϕi?

B) Determine the simple harmonic period of a point on the string.

Answer 1

Answer:

A) Φ = 0 , B)  T = 7.76 s

Explanation:

A) to find the value of the phase constant replace the value

          0 = a sin (b (0- 0) + Φ)

           0 = sin Φ

           Φ = sin⁻¹ 0

           Φ = 0

B) the period is defined by time or when the movement begins to repeat itself

So that the sine function is repeated when the angle passes 2pi

            b (x- ct) = 2pi

If we are at a fixed point x = 0

           b c t = 2pi

            t = 2π / bc

Let's calculate

            T = 2π / (33.05 245)

            T = 7.76 s

Answer 2

Answer:

A. Φi = 0

B. Simple harmonic period, T = 0.0445s

Explanation:

Parameters given:

a = 0.00580 m

b = 33.05 m-1

c = 245 m/s

A. At time t = 0 and point x = 0, f(x, t) = 0. Hence, the string is not moving at all. It is static and in its start position.

f(x, t) = asin[b(x - ct) + Φi]

f(0,0) = asin[b(0 - 0) + Φi] = 0

asin[b(0) - Φi] = 0

asin[Φi] = 0

sinΦi = 0

Φi = sin-1(0)

Φi = 0

B. By comparing the given function with the general wave function,

f(x, t) = Asin(kx - ωt + Φ)  ;  f(x, t) = asin(bx - bct + Φi)

a = A (amplitude of the wave)

b = k (wave number)

bc = ω (angular frequency)

The period of the simple harmonic motion can then be found using:

T = 2π/ω

=> T = 2π/bc

T = 2π/(33×245)

T = 0.0445 s

The simple harmonic period is 0.0445s