The vibrations of a string fixed at both ends are described by the equation $y = (5.00 m m)$ $sin [(1.57 c m^{- 1}) x] sin [(314 s^{- 1}) t]$ . What is the maximum displacement of a particle at $x = 5.66 c m$ ?

3.50 m m

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2.50 m m

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4.50 m m

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5.50 m m

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Solution

The correct option is B $2.50 m m$ .
Here,
At $x = 5.66$
Equation becomes,
$y = 5.00 sin (1.57 \times 5.66) sin (314 t)$ (in mm)

$y = 5 sin (8.8862) sin (314 t)$

$y = 5 \times 0.5 sin (314 t)$

$y = 2.50 sin (314 t)$

Maximum displacement $= 2.50 m m$

Option $B$ is the correct answer.

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Similar questions

y = (5.00 m m)

sin [(1.57 c m^{- 1}) x] sin [(314 s^{- 1}) t]

10.0 c m

y = (5.00 m m)

sin [(1.57 c m^{- 1}) x] sin [(314 s^{- 1}) t]

The vibrations of a string fixed at both ends are described by the

$y = (5.00 m m) s i n [(1.57 c m^{- 1}) x] s i n [(314 s^{- 1}) t]$

If the length of the string is 10.0 cm, locate the nodes and the antinodes. How many loops are formed in the vibration?

The vibrations of a string fixed at both ends are described by the

$y = (5.00 m m) s i n [(1.57 c m^{- 1}) x] s i n [(314 s^{- 1}) t]$

If the length of the string is 10.0 cm, locate the nodes and the antinodes. How many loops are formed in the vibration?

Two waves passing through a region are represented by

$y = 1.0 c m s i n [(3.14 c m^{- 1}) x - (157 s^{- 1}) t]$

and $y = 1.5 c m s i n [(1.57 c m^{- 1}) x - (314 s^{- 1}) t]$

Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.

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