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son4ous [18]
2 years ago
12

A point on a string undergoes simple harmonic motion as a sinusoidal wave passes. When a sinusoidal wave with speed 24 m/s, wave

length 30 cm, and amplitude of 1.0 cm passes, what is the maximum speed of a point on the string?
Physics
1 answer:
Anit [1.1K]2 years ago
6 0

Answer:

5.03 m/s

Explanation:

Give

Speed = v = 24m/s

Wavelength = λ = 30cm = 0.3m

Amplitude = A = 1.0cm = 0.01m

The velocity of a point in Simple Harmonic Motion

at any time t is given by the following formula

v = ωA cos ωt

The value is the Maximum when cosωt.

The maximum value of cosωt. is 1.

Hence the maximum velocity is ωA

Velocity of the wave v=n λ

n = v/ λ = 24 /0.3 = 80

ω = 2πn = 2π*80 = 502.86 rad/s

Maximum velocity of the particle is

ωA = 502.86 * 0.01 = 5.03m/s

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grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

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Here, The center of gravity is at its center

According to figure,

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Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

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(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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