Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto the diameter of a circle. When this is done, the analog along the diameter of the acceleration of the particle executing simple harmonic motion is

Answer:

382Hz

Explanation:

The question lacks the required option. Find the complete question in the attachment.

The long organ pipe open at both ends is called an open pipe. The fundamental frequency for an open pipe is expressed as F0 = V/2L

Harmonics are integral multiples of the fundamental frequency. For open pipes its harmonics are 2fo, 3fo, 4fo, 5fo...

Given fundamental frequency f0 to be 85 Hz, the following frequencies will be present as a standing wave;

First overtone f1 = 2fo = 2(85) = 170Hz

Second overtone f2 = 3fo = 3(85) = 255Hz

Third overtone = 4fo = 4(85) = 340Hz

Based on the option it can be seen that the only frequency that is not present as a standing wave is 382Hz

Answer:

b) the heavier one will have twice the kinetic energy of the lighter one.

Explanation:

The kinetic energy of object with mass, m

K.E₁ = ¹/₂mv²

where;

m is mass of the object

v is the velocity of the object

Since, the two objects are falling under same acceleration due to gravity, their velocity will be increasing at the same rate

The kinetic energy of object with mass, 2m

K.E₂ = ¹/₂(2m)v²

K.E₂ = 2(¹/₂mv²)

BUT K.E₁ = ¹/₂mv²

K.E₂ = 2(K.E₁)

Therefore, the heavier one will have twice the kinetic energy of the lighter one.

b) the heavier one will have twice the kinetic energy of the lighter one.

We know

[tex]\boxed{\sf m=-\dfrac{v}{u}}[/tex]

[tex]\\ \sf\longmapsto 3=-\dfrac{-40}{u}[/tex]

[tex]\\ \sf\longmapsto 3=\dfrac{40}{u}[/tex]

[tex]\\ \sf\longmapsto u=\dfrac{40}{3}[/tex]

[tex]\\ \sf\longmapsto u=13.3cm[/tex]

Answer:

Change in the frequency (in Hz) = 104.96 Hz

Explanation:

Given:

Speed of sound in air (v) = 343 m/s

Speed of car (v1) 36 m/s

Frequency(f) = 500 Hz

Find:

Change in the frequency (in Hz)

Computation:

Frequency hear by the observer(before)(f1) = [f(v+v1)] / v

Frequency hear by the observer(f1) = [500(343+36)] / 343

Frequency hear by the observer(f1) = 552.48 Hz

Frequency hear by the observer(after)(f2) = [f(v-v1)] / v

Frequency hear by the observer(f2) = [500(343-36)] / 343

Frequency hear by the observer(f2) = 447.52 Hz

Change in the frequency (in Hz) = f1 - f2

Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz

Change in the frequency (in Hz) = 104.96 Hz