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irakobra [83]
2 years ago
6

Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.2 kg attached to it and has a length of 5 m. Pendulum 2 has a ba

ll of mass 0.6 kg attached to a string of length 1 m. How does mass of the ball affect the frequency of the pendulum
Physics
2 answers:
Margarita [4]2 years ago
8 0

Answer:

Explanation:

Given:

Pendulum 1:

Mass, m = 0.2 kg

Length, l = 5 m

Pendulum 2:

Mass, m = 0.6 kg

Length, l = 1 m

Period, T = 2pi × sqrt(l/g)

Note:

Frequency, f = 1/T

Mass is not a parameter in the formula for calculating frequency of the pendulum. Therefore, mass does not affect the frequency of the motion.

Let's find the frequency of both pendulum irrespective of their masses.

T1 = 2pi × sqrt(5/9.8)

= 4.488 s

Frequency, f1 = 0.22 Hz

T2 = 2pi × sqrt(1/9.8)

= 2.007 s

Frequency, f1 = 0.498 Hz

= 0.5 Hz

mars1129 [50]2 years ago
3 0

Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

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The period of the sound wave at the given frequency is determined as 0.00235 second.

<h3>Period of the sound wave</h3>

The period of the sound wave at the given frequency is calculated as follows;

Period is reciprocal of frequency.

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A negative charge of -0.550 m exerts an upward <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="5868766e
almond37 [142]

Answer:

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Explanation:

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since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

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