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anastassius [24]
3 years ago
9

A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i

n its third OVERTONE, and the equation for the vertical displacement of any point on the string is y(x,t) = (1.22 cm) sin[(14.4 m-1)x] cos[(166 rad/s)t]. (a) What are the frequency and wavelength of the fundamental mode of this string? (b) How long is the string? (c) How fast do waves travel on this string?
Physics
1 answer:
Fofino [41]3 years ago
4 0

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
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Answer:

T = 295.57 s

Explanation:

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now, time taken by before the rocket hit the ground

using equation of motion

s = u t +\dfrac{1}{2}at^2

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negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

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