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krok68 [10]
1 year ago
11

A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered

to the string is quadrupled, by what factor does the amplitude change?.
Physics
1 answer:
marysya [2.9K]1 year ago
3 0

When the amplitude varies, it means twice as much.

<h3>Which instances of sinusoidal waves are there?</h3>

Simple harmonic motion, such the swinging of a pendulum or the weight of a weight on a spring, produces a sinusoidal relationship between position and time. Sound and water waves, for example, can be visualized as sinusoids.

Given:

if the string is given four times as much power.

According to the power that a sinusoidal wave on a stretched string transmits,

P =\frac{1}{2} μω^A^2v

Where  μ=mass per unit length, w= angular speed, A= amplitude, v= wave speed, P= power delivered.

P ∝A^{2}

Here, the only terms we need are power and amplitude.

A =K\sqrt{P}

Where K is constant

Here, "power become quadruple" refers to four,

A = \sqrt{4P}

A =2 \sqrt{P}

Therefore, an increase in amplitude by a factor of two is equivalent to a twofold increase.

To know more about sinusoidal wave visit:-

brainly.com/question/28449631

#SPJ4

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If the mass of the ladder is 12.0 kg, the mass of the painter is 55.0 kg, and the ladder begins to slip at its base when her fee
arsen [322]

Answer:

μ = 0.498

Explanation:

I manage to find the picture of this problem. In the attached picture you can also see the forces involved in this case.

We know according to newton's law that the static friction force is:

Fs = μ * N

However, as you can see in the picture, the Normal force is equals to the weight, in this case the weight of the painter and also the ladder, therefore:

N = Fy

and the force of the friction is:

Fs = Fx

Therefore the coefficient is:

μ = Fx/Fy  (1)

Now, let's write the equations in x and y, to solve this.

For the "x" axis:

Fx - Fw = 0 -----> Fx = Fw   (2)

Fw is force of the wall, while Fx is the force friction in the x axis (base of the ladder).

for the "y" axis:

Fy - W1 - W2 = 0

W1 = mg (ladder)

W2 = Mg (painter)

replacing we have:

Fy = W1 + W2

Fy = mg + Mg ----> Fy = g(m + M)    (3)

To get the force that the wall is exerting we need to calculate the torque around the foot of the ladder so:

τ = rF sinθ

However, the angle in the wall and the ladder is 90° so:

τ = rF sin90°

τ = rF   (4)

replacing (4) with the forces we have:

rFw = rW1 + rW2

4Fw = 1.5mg + 2.1Mg

Fw = 1.5mg + 2.1Mg/4   (5)

Finally, with this expression, we can replace it in (1) to get the coefficient of friction:

μ = Fx/Fy

μ = 1.5mg + 2.1Mg / 4g(m + M)     gravity cancels out so:

μ = 1.5m + 2.1M / 4(m + M)

Replacing the data we finally have:

μ = (1.5 * 2) + (2.1 * 55) / 4 (55 + 12)

μ = 133.5 / 268

<h2>μ = 0.498</h2>

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Answer: 7

Explanation:

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