A particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

3 answers
Question:

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre of its motion​(b) the maximium speed (c) the maximium acceleration

Answers

a) [tex]speed=\pi cm/s[/tex]

b) [tex]v_{max}=\frac{5\pi}{4} cm/s[/tex]

c) [tex]a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}[/tex]

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

[tex]y=Asin(\omega t)[/tex]

where:

A=amplitude

[tex]\omega[/tex]= angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

[tex]\omega=\frac{2\pi}{T}[/tex]

so our angular frequency is:

[tex]\omega=\frac{2\pi}{8s}[/tex]

[tex]\omega=\frac{\pi}{4}[/tex]

so now we can build our equation:

[tex]y=5sin(\frac{\pi}{4} t)[/tex]

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

[tex]y=5sin(\frac{\pi}{4} t)[/tex]

[tex]3=5sin(\frac{\pi}{4} t)[/tex]

so we solve for t:

[tex]sin(\frac{\pi}{4} t)=\frac{3}{5}[/tex]

[tex]\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})[/tex]

[tex]t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})[/tex]

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

[tex]y=5sin(\frac{\pi}{4} t)[/tex]

[tex]y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}[/tex]

which simplifies to:

[tex]y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)[/tex]

and we can now substitute the t-value we found previously, so we get:

[tex]y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))[/tex]

[tex]y'=\pi[/tex]

so its velocity at that point is [tex]\pi[/tex] cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

[tex]y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)[/tex]

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

[tex]A=\frac{5\pi}{4}[/tex]

so:

[tex]v_{max}=\frac{5\pi}{4} cm/s[/tex]

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

[tex]y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)[/tex]

We can use the chain rule again to find this derivative so we get:

[tex]y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})[/tex]

so when simplified we get:

[tex]y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)[/tex]

its amplitude is:

[tex]A=\frac{5\pi^{2}}{16}[/tex]

so its maximum acceleration is:

[tex]a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}[/tex]

sonar - device which is used to find the submit submarines, sankalp ship. sonar waves consists of a transmitter and detector which are installed in the observation centre on board of a ship. the ultrasonic waves are sent into the sea at high frequency about 1000 khz in all direction.

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

[tex]Look at the pendulum diagram between which points does the pendulum’s potential energy increase most[/tex]

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