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lesantik [10]
3 years ago
5

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?
Physics
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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Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

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A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is
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Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

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