A string, 2 m long and fixed at both ends, is set into vibrations in its first overtone. The wave speed on the string is 200 m s^ -1 and the amplitude is 0.5 cm . Determine the equation giving the displacement of

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Correct answer: D. (0.5 cm) [(π m^ -1 )x] [(200π s^ -1 )t]

For a string fixed at both ends, the wavelength in the first overtone (

n = 2

) is given by

L = 2 \left(\dfrac{\lambda}{2}\right) = \lambda

.

Given

L = 2 \text{ m}

, we have

\lambda = 2 \text{ m}

.

The wave number

k

is

k = \dfrac{2\pi}{\lambda} = \pi \text{ m}^{-1}

.

The angular frequency

\omega

is

\omega = v k = 200 \times \pi = 200\pi \text{ s}^{-1}

.

The general equation for a standing wave with nodes at

x = 0

and

x = L

is

y(x, t) = A \sin(kx) \cos(\omega t + \phi)

.

Given the amplitude

A = 0.5 \text{ cm}

.

At

t = 0

, the point

x = 50 \text{ cm} = 0.5 \text{ m}

has maximum displacement.

Substituting

x = 0.5 \text{ m}

into the spatial part gives

\sin(\pi \times 0.5) = 1

, which means it is an antinode.

For the displacement to be maximum at

t = 0

, the temporal part

\cos(\omega \times 0 + \phi) = \cos(\phi)

must be

1

, giving

\phi = 0

.

Therefore, the equation of the standing wave is

y(x, t) = (0.5 \text{ cm}) \sin[(\pi \text{ m}^{-1})x] \cos[(200\pi \text{ s}^{-1})t]

.

A string, 2 m long and fixed at both ends, is set into vibrations...

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