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tankabanditka [31]
2 years ago
15

A slinky spring 3m long rests on a horizontal bench with one end fixed. When the free end is suddenly pushed forward, the compre

ssion pulse travels to the fixed end and back in 2 secs. What is the speed of the longitudinal waves along the spring? What is the wavelength of the longitudinal waves produced when the free end is moved to and fro with a frequency of 2.5Hz? A slinky spring 3m long rests on a horizontal bench with one end fixed . When the free end is suddenly pushed forward , the compression pulse travels to the fixed end and back in 2 secs . What is the speed of the longitudinal waves along the spring ? What is the wavelength of the longitudinal waves produced when the free end is moved to and fro with a frequency of 2.5Hz ?​
Physics
1 answer:
Papessa [141]2 years ago
6 0

The speed of the longitudinal waves along the spring is 1.5 m/s.

The wavelength of the longitudinal waves produced when the free end is moved to and fro is 0.6 m.

<h3>What is wavelength?</h3>

The wavelength is the distance between the adjacent crest or trough of the sinusoidal wave. The wavelength is the reciprocal of the frequency of the wave.

Wavelength λ = c/f

where c is the speed of light in vacuum = 3*10⁸ m/s

A slinky spring 3m long rests on a horizontal bench with one end fixed. When the free end is suddenly pushed forward, the compression pulse travels to the fixed end and back in 2 secs.

λ = v/f

v =fλ =λ /T where T is the time period = 2sec

Put the values we get

v = 3/2 =1.5 m/s

Thus, the speed of wave is 1.5 m/s.

Using the same expression for frequency 2.5Hz, wavelength is

λ = v/f

λ = 1.5 /2.5 = 0.6 m

Thus,  the wavelength is 0.6m.

Learn more about wavelength.

brainly.com/question/13533093

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A ball, with a mass of 4.5kg, is thrown directly upwards. It reaches a maximum height of 19m from the point at which it was rele
pishuonlain [190]

According to the principle of energy conservation, <em><u>the energy is not created, nor destroyed, it is transformed.</u></em>


In this problem, we are talking about Mechanical Energy (M) which is the addition of the Kinetic Energy K (energy of the body in motion) and Potential Energy P (It can be Gravitational Potential Energy or Elastic Potential Energy, in this case is the first one):



M=K+P   (1)



The Kinetic Energy is: K=\frac{1}{2}mV^{2}


Where m is the mass of the body and V its velocity



And the Potential Energy (Gravitational) is: P=mgh


Where g is the gravitational acceleration and h is the height of the body.



Knowing this, the equation for the Mechanical Energy in this case is:


M=\frac{1}{2}mV^{2}+mgh   (2)



Now, according to the <u>Conservation of the Energy Principle</u>, the initial energy M_{i} must be equal to the final energy M_{f}:


M_{i}=M_{f}    (3)



M_{i}=\frac{1}{2}m{V_{i}}^{2}+mgh_{i}    (4)



At the beginning, the ball is in h_{i}=0m over the point where it is released and has an initial speed  V_{i}, this means the initial energy M_{i} is only the Kinetic Energy:



M_{i}=\frac{1}{2}m{V_{i}}^{2}    (5)



At the maximum height of h_{f}=19m from the point at which the ball was released, it has an speed V_{f}=0\frac{m}{s}, <u>because at that very moment the ball stops and then begins to fall. </u>

This means the final energy M_{f} is only the Potential Gravitational Energy



M_{f}=mgh_{f}    (6)



Well, according to the explanation above, we have to substitute (5) and (6) in (3):



\frac{1}{2}m{V_{i}}^{2}=mgh_{f}    (7)



Now we have to find V_{i}, the velocity of the ball when it was released:



\frac{1}{2}(4.5kg){V_{i}}^{2}=(4.5kg)(9.8\frac{m}{s^2})(19m)    


4.5kg{V_{i}}^{2}=2(837.9\frac{kgm^2}{s^2})    


{V_{i}}^{2}=\frac{1675.8\frac{kgm^2}{s^2}}{4.5kg}    


{V_{i}}=\sqrt{372.4\frac{m^2}{s^2}}    


Finally:


{V_{i}}=19.297\frac{m}{s}    


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3 years ago
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