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enyata [817]
2 years ago
5

A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)​

Physics
1 answer:
Vadim26 [7]2 years ago
7 0

Answer:

\huge\boxed{\sf f=0.55 \ Hz}

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }

<u>Solution:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}

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The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre
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Answer:

a) P_m=1.78\ Pa

b) f=79.5775\ Hz

c) \lambda=7.076\ m

d) v=563.06\ m.s^{-1}

Explanation:

<u>Given equation of pressure variation:</u>

\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]

We have the standard equation of periodic oscillations:

\Delta P=P_m\ sin\ (kx-\omega.t)

<em>By comparing, we deduce:</em>

(a)

amplitude:

P_m=1.78\ Pa

(b)

angular frequency:

\omega=2\pi.f

2\pi.f=500

∴Frequency of oscillations:

f=\frac{500}{2\pi}

f=79.5775\ Hz

(c)

wavelength is given by:

\lambda=\frac{2\pi}{k}

\lambda=\frac{2\pi}{0.888}

\lambda=7.076\ m

(d)

Speed of the wave is gives by:

v=\frac{\omega}{k}

v=\frac{500}{0.888}

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8 0
3 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
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