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3 years ago

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1. A pendulum has a period of 3 seconds. What's its frequency? 2. A pendulum has a frequency of 0.25 Hz. What's its period? 3. A

pendulum makes 10 back and forth swings in 20 seconds. What's its period AND frequency?

2 answers:

RideAnS [48]3 years ago

6 0

Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

(1) We have given time period of pendulum is 3 sec

So T = 3 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{3}=0.333Hz$

(2) Frequency of the pendulum is given f = 0.25 Hz

Time period is equal to $T=\frac{1}{f}=\frac{1}{0.25}=4sec$

(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds

So time taken to complete 1 back and forth swings = $=\frac{20}{10}=2sec$

So time period T = 2 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{2}=0.5Hz$

andre [41]3 years ago

6 0

Answer:

(1) 0.33 Hz

(2) 4 s

(3) 2 s, 0.5 Hz

Explanation:

The relation between frequency and time period is given as

$f=\frac{1}{T}$

(1) Given a pendulum has a period, T = 3 s.

So, $f=\frac{1}{3s}$

f = 0.333 Hz.

Thus, frequency of pendulum is 0.33 Hz.

(2) Given a pendulum has a frequency, f = 0.25 Hz.

So, $T=\frac{1}{0.25 Hz}$

T = 4 s.

Thus , period of pendulum is 4 s.

(3) Frequency is measured in cycles/second.

Given a pendulum makes 10 back and forth swings in 20 seconds.

So, $f=\frac{10 cycles}{20s}$

f = 0.5 Hz

Time period,

$T=\frac{1}{0.5}$

T = 2 s.

Thus, period and frequency of pendulum are 2 s and 0. 5 Hz respectively.

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3 years ago

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Check the explanation section for a better understanding

Explanation:

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$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

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ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

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You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

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d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

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3 0

3 years ago