two pendula are released from rest at the same height h above their lowest points. how do the maximum velocities of the two pendulums compare? note that pendulum 2 has a string with length twice that of pendulum 1.

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Two simple pendulums are released from rest a height h above the lowest point. How do the maximum velocities of the two pendulums compare? Pendulum 2 has a length twice the length of pendulum 1. a. v

Answer to: Two simple pendulums are released from rest a height h above the lowest point. How do the maximum velocities of the two pendulums...

Pendulum

Two simple pendulums are released from rest a height h above the lowest point. How do the maximum...

Two simple pendulums are released from rest a height h above the lowest point. How do the maximum... Question:

Two simple pendulums are released from rest a height h above the lowest point. How do the maximum velocities of the two pendulums compare? Pendulum 2 has a length twice the length of pendulum 1.

<v1,max

a. v 1 , m a x > v 2 , m a x v1,max>v2,max b. v 1 , m a x v 2 , m a x

c. v 1 , m a x = v 2 , m a x v1,max=v2,max

d. you cannot determine the relative speeds without numbers.

Maximum Velocity of the Simple Pendulum at the Lowest Point

Velocity of the simple pendulum at the lowest point depends on the maximum potential energy of the bob at its highest position. This potential energy of the bob of the pendulum at the highest point will be converted into kinetic energy of the bob of the pendulum at the lowest point. In order to calculate the potential energy of the pendulum normally the height has to be expressed in terms of the length of the pendulum and angle subtended by the pendulum with the vertical.

Answer and Explanation:

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Given points

Height of the maximum position for both pendulums from the mean position

Length of pendulum one

L 1 L1

Length of pendulum...

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Pendulums in Physics: Definition & Equations

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Chapter 11 / Lesson 6

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Understand the definition of a pendulum in physics. Learn how Newtonian mechanics describes the motion of pendulums, their period and frequency, through equations.

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15.4 Pendulums

15.4 Pendulums

LEARNING OBJECTIVES

By the end of this section, you will be able to:

State the forces that act on a simple pendulum

Determine the angular frequency, frequency, and period of a simple pendulum in terms of the length of the pendulum and the acceleration due to gravity

Define the period for a physical pendulum

Define the period for a torsional pendulum

Pendulums are in common usage. Grandfather clocks use a pendulum to keep time and a pendulum can be used to measure the acceleration due to gravity. For small displacements, a pendulum is a simple harmonic oscillator.

The Simple Pendulum

A simple pendulum is defined to have a point mass, also known as the pendulum bob, which is suspended from a string of length L with negligible mass ((Figure)). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string. The mass of the string is assumed to be negligible as compared to the mass of the bob.

Figure 15.20 A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of

− m g sin θ −mgsinθ

toward the equilibrium position—that is, a restoring force.

Consider the torque on the pendulum. The force providing the restoring torque is the component of the weight of the pendulum bob that acts along the arc length. The torque is the length of the string L times the component of the net force that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the opposite direction of the angular displacement:

τ = − L ( m g sin θ ) ; I α = − L ( m g sin θ ) ; I d 2 θ d t 2 = − L ( m g sin θ ) ; m L 2 d 2 θ d t 2 = − L ( m g sin θ ) ; d 2 θ d t 2 = − g L sin θ .

τ=−L(mgsinθ);Iα=−L(mgsinθ);Id2θdt2=−L(mgsinθ);mL2d2θdt2=−L(mgsinθ);d2θdt2=−gLsinθ.

The solution to this differential equation involves advanced calculus, and is beyond the scope of this text. But note that for small angles (less than 15 degrees),

sin θ sinθ and θ θ

differ by less than 1%, so we can use the small angle approximation

sin θ ≈ θ . sinθ≈θ. The angle θ θ

describes the position of the pendulum. Using the small angle approximation gives an approximate solution for small angles,

d 2 θ d t 2 = − g L θ . d2θdt2=−gLθ.

Because this equation has the same form as the equation for SHM, the solution is easy to find. The angular frequency is

ω = √ g L ω=gL and the period is T = 2 π √ L g . T=2πLg.

The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if

θ θ is less than about 15 ° . 15°.

Even simple pendulum clocks can be finely adjusted and remain accurate.

Note the dependence of T on g. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity, as in the following example.

EXAMPLE

Measuring Acceleration due to Gravity by the Period of a Pendulum

What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?

Strategy

We are asked to find g given the period T and the length L of a pendulum. We can solve

T = 2 π √ L g T=2πLg

for g, assuming only that the angle of deflection is less than

15 ° 15° .

Solution

Square T = 2 π √ L g T=2πLg and solve for g: g = 4 π 2 L T 2 . g=4π2LT2.

Substitute known values into the new equation:

g = 4 π 2 0.75000 m ( 1.7357 s ) 2 .

g=4π20.75000m(1.7357s)2.

Calculate to find g:

g = 9.8281 m/s 2 . g=9.8281m/s2.

Significance

This method for determining g can be very accurate, which is why length and period are given to five digits in this example. For the precision of the approximation

Source : courses.lumenlearning.com

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AP Physics 1 Help » Newtonian Mechanics » Circular, Rotational, and Harmonic Motion » Harmonic Motion » Pendulums

Example Question #1 : Pendulums

A ball of mass 2kg is attached to a string of length 4m, forming a pendulum. If the string is raised to have an angle of 30 degrees below the horizontal and released, what is the velocity of the ball as it passes through its lowest point?

Possible Answers:

4.2 m s 2.1 m s 3.2 m s 6.3 m s 0 m s

Correct answer:

6.3 m s Explanation:

This question deals with conservation of energy in the form of a pendulum. The equation for conservation of energy is:

E= U i + K i = U f + K f

According to the problem statement, there is no initial kinetic energy and no final potential energy. The equation becomes:

U i = K f

Subsituting in the expressions for potential and kinetic energy, we get:

mgh= 1 2 m v 2 f

We can eliminate mass to get:

gh= 1 2 v 2 f

Rearranging for final velocity, we get:

v f = 2gh − − − √

In order to solve for the velocity, we need to find the initial height of the ball.

The following diagram will help visualize the system:

From this, we can write:

4=d+h h=4−d

Using the length of string and the angle it's held at, we can solve for

d : d=4sin( 30 ∘ )=2 h=4−d=4−2=2

Now that we have all of our information, we can solve for the final velocity:

v f = 2⋅10⋅2 − − − − − − − √ = 40 − − √ =6.3 m s Report an Error

Example Question #1 : Pendulums

A pendulum has a period of 5 seconds. If the length of the string of the pendulum is quadrupled, what is the new period of the pendulum?

g=10 m s 2

Possible Answers:

2.5s 1.25s 20s 5s 10s

Correct answer:

10s Explanation:

We need to know how to calculate the period of a pendulum to solve this problem. The formula for period is:

P=2π L g − − √

In the problem, we are only changing the length of the string. Therefore, we can rewrite the equation for each scenario:

P 1 = 2π g √ L 1 − − √ P 2 = 2π g √ L 2 − − √

Dividng one expression by the other, we get a ratio:

P 1 P 2 = L 1 L 2 − − − √ We know that L 2 =4 L 1

, so we can rewrite the expression as:

P 1 P 2 = L 1 4 L 1 − − − − √ = 1 4 − − √ = 1 2

Rearranging for P2, we get:

P 2 =2 P 1 =2(5s)=10s Report an Error

Example Question #1 : Pendulums

A student studying Newtonian mechanics in the 19th century was skeptical of some of Newton's concepts. The student has a pendulum that has a period of 3 seconds while sitting on his desk. He attaches the pendulum to a ballon and drops it off the roof of a university building, which is 20m tall. Another student realizes that the pendulum strikes the ground with a velocity of

12 m s

. What is the period of the pendulum as it is falling to the ground?

Neglect air resistance and assume

g=10 m s 2

Possible Answers:

4s 1s 5s 2s 3s

Correct answer:

5s Explanation:

We need to know the formula for the period of a pendulum to solve this problem:

T=2π L g − − √

We aren't given the length of the pendulum, but that's ok. We could solve for it, but it would be an unnecessary step since the length remains constant.

We can write this formula for the pendulum when it is on the student's table and when it is falling:

T 1 =2π L g 1 − − − √ T 2 =2π L g 2 − − − √

1 denotes on the table and 2 denotes falling. The only thing that is different between the two states is the period and the gravity (technically the acceleration of the whole system, but this is the form in which you are most likely to see the formula). We can divide the two expressions to get a ratio:

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