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3 years ago

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A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the ve

ry small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)Assume that the tension of the string is constant and equal to W. (a) How much time does it take a pulse to travel the fulllength of the string? (b) What is the weight W? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string?

1 answer:

Natalka [10]3 years ago

5 0

Answer:

Explanation:

mass of string = .0125 / 9.8

= 1.275 x 10⁻³ kg

Length of string l = 1.5 m .

m = mass per unit length

= ( .1.275 / 1.5) x 10⁻³ kg/m

m = .85 x 10⁻³ kg/m

wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

compare with equation of wave

y(x,t) = Acos(K x − ω t)

ω ( angular velocity ) = 4830 rad/s

k = 172 rad/m

Velocity = ω / k

= 4830/172 m /s

= 28.08 m /s

velocity of wave = $\sqrt{\frac{W}{m } }$

28.08 = $\sqrt{\frac{W}{.85\times10^{-3} } }$

788.48 = W / .85 X 10⁻³

W = 670 x 10⁻³ N .

c ) wave length

wave length =2π / k

= 2 x 3.14 / 172

= .0365 m

no of wave lengths over whole length of string

= 1.5 / .0365

= 41

d )

equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

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Answer:

1. 12 cm

2. 0.133 m

3. 0.03 m

4. Plane mirror

Virtual image

Upright

Behind the mirror

The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Real image

Inverted image

In front of the the mirror

Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Virtual image

Upright image

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Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object

Explanation:

1. For plane mirror, since there is no magnification, the virtual image distance from the mirror = object distance from the mirror = 12 cm behind the mirror

2. The height of the object = 0.3 m

The distance of the object from the mirror = 0.4 meters

Height of image formed = 0.1 meter

We have;

$Magnification, \ m = \dfrac{Image \ height }{Object \ height } = \dfrac{Image \ distance \ from \ mirror }{Object\ distance \ from \ mirror }$

$m = \dfrac{0.1}{0.3 } = \dfrac{Image \ distance \ from \ mirror }{0.4 }$

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Image distance from the mirror = 0.133 m

3. $m = \dfrac{Image \ height}{0.10 } = \dfrac{0.06 }{0.20 }$

The image height = 0.06/0.2×0.1 = 3/100 = 0.03 meter

The image height = 0.03 meter

4. Plane mirror

Type = Virtual image

Appearance = Upright image with the left transformed to right

Placement = Behind the mirror

Size = The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Type = Real image

Appearance = Inverted image

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Size = Diminished when the object is beyond the center of curvature

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Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

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Answer:

Part a)

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Part b)

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Explanation:

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$F_{net} = F_x - F_k$

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$a = \frac{F_{net}}{m}$

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Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

$F_x = 72.5 N$

$F_y = 33.8 N$

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

$F_n = (25)(9.81)cos10 - 33.8 = 207.7 N$

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$f_k = \mu F_n$

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$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

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Answer:

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