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ValentinkaMS [17]
3 years ago
10

A simple harmonic transverse wave is propagating along a string towards the left direction as shown in the figure. figure shows

a plot of displacement as function of position at time t=0.The string tension is 3.6 Newton and it’s linear density is 25GM/M. Calculate
I) the amplitude
Ii) the wavelength
Iii) wave speed
Iv) the period
V) The maximum particle speed in the String
Physics
1 answer:
dimulka [17.4K]3 years ago
6 0

Answer:

  1. Amplitude  = 5 cm (Heights)
  2. Wavelength (λ)  = 40 cm
  3. Wave speed (v) = 12 m/s (Approx)
  4. Time period (T) = 0.033 s (Approx)
  5. Maximum particle speed (V) = 9.43 m/s

Explanation:

1) Amplitude

Amplitude  = 5 cm (Heights)

2) Wavelength (λ)

Wavelength (λ)  = 40 cm

3) Wave speed

Wave speed (v) = √ t / μ

Wave speed (v) = √ 3.6 / [25x10⁻³]

Wave speed (v) = 12 m/s (Approx)

4) Time period (T)

Time period (T) = 1/f = (λ)/v

Time period (T) = 0.40m / 12

Time period (T) = 0.033 s (Approx)

5) Maximum particle speed (V)

Maximum particle speed (V) = Aw

Maximum particle speed (V) = [0.05x2x3.14] / 0.033

Maximum particle speed (V) = 9.43 m/s

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Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

a= \frac{v^2}{r}

And the velocity is given by:

v=\frac{2\pi r}{T}

If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

And we can replace the values and we got:

a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
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Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

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4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

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c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

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