Question 1

An astronaut has landed on an asteroid and conducts an experiment to determine the acceleration of gravity on that asteroid.He uses a simple pendulum that has a period of oscillation of 2.00 s on Earth and finds that on the asteroid the period is 11.3 s.What is the acceleration of gravity on that asteroid?

Accepted Answer

The period of a pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$, where $T$ is the period, $L$ is the length of the pendulum, and $g$ is the acceleration due to gravity. Since the length of the pendulum doesn't change, the ratio of the squares of the periods on Earth and the asteroid gives the inverse ratio of the gravitational accelerations. Thus, $\left(\frac{T_{asteroid}}{T_{Earth}}\right)^2 = \frac{g_{Earth}}{g_{asteroid}}$. Given $T_{Earth} = 2.00s$, $T_{asteroid} = 11.3s$, and $g_{Earth} = 9.81 m/s^2$, solving for $g_{asteroid}$ yields approximately $0.307 m/s^2$.

Question 2

On the Moon,the acceleration of gravity is g/6.If a pendulum has a period T on Earth,what will its period be on the Moon?

Accepted Answer

The answer of On the Moon,the acceleration of gravity is...

Question 3

The period of a simple pendulum that is 1.00 m long on another planet is 
What is the acceleration due to gravity on this planet if the mass of the pendulum bob is 1.5 kg?

Accepted Answer

The answer of The period of a simple pendulum that...

Question 4

What is the length of a simple pendulum with a period of 2.0 s?

Accepted Answer

The period of a simple pendulum is given by:

T = 2π√(L/g)

where L is the length of the pendulum and g is the acceleration due to gravity (9.81 m/s²).

Rearranging this equation, we get:

L = (gT²)/(4π²)

Substituting the given values, we get:

L = (9.81 m/s²)(2.0 s)²/(4π²) ≈ 0.99 m

Therefore, the best choice is B.

Question 5

A 0.50-kg object is attached to an ideal spring of spring constant (force constant)20 N/m along a horizontal,frictionless surface.The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.At what distance from the equilibrium position are the kinetic energy and potential energy of the system the same?

Accepted Answer

The answer of A 0.50-kg object is attached to an...

Question 6

A ball vibrates back and forth from the free end of an ideal spring having a force constant (spring constant)of 20 N/m.If the amplitude of this motion is 0.30 m,what is the kinetic energy of the ball when it is 0.30 m from its equilibrium position?

Accepted Answer

At 0.30 m from its equilibrium position, the ball is at its maximum displacement (amplitude), where all the energy is potential and the kinetic energy is zero.

Question 7

A Christmas ornament made from a thin hollow glass sphere hangs from a small hook at its surface.It is observed to oscillate with a frequency of 2.50 Hz in a city where g = 9.80 m/s².What is the radius of the ornament? The moment of inertia of a hollow sphere of mass M and radius R about a point on its edge is 5MR²/3.

Accepted Answer

We can use the formula for the frequency of oscillation of a simple pendulum:

f = 1/2π √(g/L)

where g is the acceleration due to gravity and L is the length of the pendulum. In this case, the glass sphere is acting as a simple pendulum, and its length L is equal to the radius of the sphere, r.

Rearranging the formula, we get:

r = g/4π^2 f^2

Plugging in the given values, we get:

r = 9.80 m/s^2 / (4π^2 x 2.50 Hz)^2
r = 2.38 cm

Finally, we can check that our answer is reasonable by comparing it to the dimensions of typical Christmas ornaments. A radius of around 2.38 cm seems reasonable for a small glass ornament. Therefore, the best choice is (C).

Question 8

A simple pendulum takes 2.00 s to make one compete swing.If we now triple the length,how long will it take for one complete swing?

Accepted Answer

The period of a pendulum is given by T=2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity. When we triple the length of the pendulum, the new length becomes 3l. Therefore, the new period T' is given by T'=2π√(3l/g)=(2π/√g)√(3l). Since g is constant, the ratio of T' to T is √(3l)/√l=√3. Thus, T'=√3×T=√3×2.00 s=3.46 s.

Question 9

An astronaut has landed on Planet N-40 and conducts an experiment to determine the acceleration due to gravity on that planet.She uses a simple pendulum that is 0.640 m long and measures that 10 complete oscillations 26.0 s.What is the acceleration of gravity on Planet N-40?

Accepted Answer

The period of a pendulum is given by T = 2π√(L/g). For 10 oscillations, the total time is 26.0 s, so the period T = 26.0 s / 10 = 2.6 s. Rearranging the formula to solve for g gives g = 4π²L/T². Substituting L = 0.640 m and T = 2.6 s, we find g ≈ 3.74 m/s².

Question 10

A 1.53-kg piece of iron is hung by a vertical ideal spring.When perturbed slightly,the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm.If we choose the total potential energy (elastic and gravitational)to be zero at the equilibrium position of the hanging iron,what is the total mechanical energy of the system?

Accepted Answer

The total mechanical energy in simple harmonic motion is given by $ E = \frac{1}{2} k A^2 $, where $ k $ is the spring constant and $ A $ is the amplitude. The spring constant $ k $ can be found using $ k = (2\pi f)^2 m $, where $ f $ is the frequency and $ m $ is the mass. Substituting the given values, we find $ k \approx 147.9 \, \text{N/m} $. Then, $ E = \frac{1}{2} \times 147.9 \times (0.075)^2 \approx 0.646 \, \text{J} $.

Question 11

A simple pendulum having a bob of mass M has a period T.If you double M but change nothing else,what would be the new period?

Accepted Answer

The answer of A simple pendulum having a bob of...

Question 12

A thin hoop is supported in a vertical plane by a nail.What should the radius of the hoop be in order for it to have a period of oscillation of 1.00 s? The moment of inertia of a hoop of mass M and radius R about a point on its rim is 2MR².

Accepted Answer

The period of oscillation for a hoop rotating about a fixed axis perpendicular to the plane of the hoop is given by:

T = 2π√(I/mgd)

where I is the moment of inertia of the hoop, m is its mass, g is acceleration due to gravity, and d is the distance between the axis of rotation and the center of mass of the hoop.

In this case, the hoop is supported by a nail, which means that the axis of rotation passes through its center of mass. Therefore, d = R, where R is the radius of the hoop.

Rearranging the equation for T, we get:

I = (mgR^2)/(4π^2T^2)

Substituting the given values, we get:

I = (2MR^2)/5

Setting T = 1.00 s, we can solve for R:

R = √[(5IT^2)/(2M)]

Substituting the given values, we get:

R = √[(5)(1.00 s)^2/(2)(M)] = √(5/2)R

Solving for R, we get:

R = 0.124 m

Therefore, the best choice is C.

Question 13

A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant)of 3.58 N/m undergoes simple harmonic oscillations with an amplitude of 7.50 cm.What is the total mechanical energy of the system?

Accepted Answer

The answer of A 0.150-kg cart that is attached to...

Question 14

A 34-kg child on an 18-kg swing set swings back and forth through small angles.If the length of the very light supporting cables for the swing is 4.9m. 
How long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables.

Accepted Answer

The period of a pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Plugging in the given length of $4.9 \, \text{m}$ for $L$, we get $T = 2\pi\sqrt{\frac{4.9}{9.8}} = 2\pi\sqrt{0.5} = 2\pi\sqrt{\frac{1}{2}}$. This simplifies to $T = 2\pi\left(\frac{1}{\sqrt{2}}\right) = \pi\sqrt{2}$. Numerically, this evaluates to approximately $4.4$ seconds, making choice A correct. The mass of the child and swing set does not affect the period of a pendulum for small angles.

Question 15

A simple pendulum has a period T on Earth.If it were used on Planet X,where the acceleration due to gravity is 3 times what it is on Earth,what would its period be?

Accepted Answer

The answer of A simple pendulum has a period T...

Question 16

In 1851 Jean Bernard Leon Foucault demonstrated the rotation of the earth using a pendulum 11.0 m long,which was set up in the Paris Observatory.How long would it have taken for Foucault's pendulum to make one complete swing back to its starting point if g = 9.81 m/s² at the observatory?

Accepted Answer

The period of a pendulum is given by the formula $T = 2\pi\sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity. Substituting $L = 11.0 \, \text{m}$ and $g = 9.81 \, \text{m/s}^2$ into the formula gives $T = 2\pi\sqrt{\frac{11.0}{9.81}} \approx 6.65 \, \text{s}$.

Question 17

A pendulum that was originally erected by Foucault at the Pantheon in Paris for the Paris Exhibition in 1851 was restored in 1995.It has a 28.0-kg sphere suspended from a 67.0-m light cable.How long would it take for the bob in this pendulum to move from the position of maximum displacement down to the equilibrium point?

Accepted Answer

The period of a pendulum is given by:$$T = 2\pi \sqrt{\frac{L}{g}}$$where L is the length of the pendulum and g is the acceleration due to gravity.In this case, L = 67.0 m and g = 9.81 m/s^2.So, the period of the pendulum is:$$T = 2\pi \sqrt{\frac{67.0 	ext{ m}}{9.81 	ext{ m/s}^2}} = 16.4 	ext{ s}$$The time it takes for the bob to move from the position of maximum displacement down to the equilibrium point is half of the period, or:$$t = \frac{T}{2} = \frac{16.4 	ext{ s}}{2} = 8.22 	ext{ s}$$So, the correct answer is A) 4.11 s.

Question 18

If both the mass of a simple pendulum and its length are doubled,the period will

Accepted Answer

The answer of If both the mass of a simple...

Question 19

How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will oscillate at 1.00 Hz?

Accepted Answer

The answer of How much mass should be attached to...

Quiz 13: Oscillations About Equilibrium

On the Moon,the acceleration of gravity is g/6.If a pendulum has a period T on Earth,what will its period be on the Moon?

The period of a simple pendulum that is 1.00 m long on another planet is What is the acceleration due to gravity on this planet if the mass of the pendulum bob is 1.5 kg?

What is the length of a simple pendulum with a period of 2.0 s?

A ball vibrates back and forth from the free end of an ideal spring having a force constant (spring constant) of 20 N/m.If the amplitude of this motion is 0.30 m,what is the kinetic energy of the ball when it is 0.30 m from its equilibrium position?

A simple pendulum takes 2.00 s to make one compete swing.If we now triple the length,how long will it take for one complete swing?

An astronaut has landed on Planet N-40 and conducts an experiment to determine the acceleration due to gravity on that planet.She uses a simple pendulum that is 0.640 m long and measures that 10 complete oscillations 26.0 s.What is the acceleration of gravity on Planet N-40?

A simple pendulum having a bob of mass M has a period T.If you double M but change nothing else,what would be the new period?

A thin hoop is supported in a vertical plane by a nail.What should the radius of the hoop be in order for it to have a period of oscillation of 1.00 s? The moment of inertia of a hoop of mass M and radius R about a point on its rim is 2MR².

A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m undergoes simple harmonic oscillations with an amplitude of 7.50 cm.What is the total mechanical energy of the system?

A simple pendulum has a period T on Earth.If it were used on Planet X,where the acceleration due to gravity is 3 times what it is on Earth,what would its period be?

In 1851 Jean Bernard Leon Foucault demonstrated the rotation of the earth using a pendulum 11.0 m long,which was set up in the Paris Observatory.How long would it have taken for Foucault's pendulum to make one complete swing back to its starting point if g = 9.81 m/s² at the observatory?

If both the mass of a simple pendulum and its length are doubled,the period will

How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 N/m so that it will oscillate at 1.00 Hz?

Quiz 13: Oscillations About Equilibrium

On the Moon,the acceleration of gravity is g/6.If a pendulum has a period T on Earth,what will its period be on the Moon?

The period of a simple pendulum that is 1.00 m long on another planet is What is the acceleration due to gravity on this planet if the mass of the pendulum bob is 1.5 kg?

What is the length of a simple pendulum with a period of 2.0 s?

A ball vibrates back and forth from the free end of an ideal spring having a force constant (spring constant) of 20 N/m.If the amplitude of this motion is 0.30 m,what is the kinetic energy of the ball when it is 0.30 m from its equilibrium position?

A simple pendulum takes 2.00 s to make one compete swing.If we now triple the length,how long will it take for one complete swing?

An astronaut has landed on Planet N-40 and conducts an experiment to determine the acceleration due to gravity on that planet.She uses a simple pendulum that is 0.640 m long and measures that 10 complete oscillations 26.0 s.What is the acceleration of gravity on Planet N-40?

A simple pendulum having a bob of mass M has a period T.If you double M but change nothing else,what would be the new period?

A thin hoop is supported in a vertical plane by a nail.What should the radius of the hoop be in order for it to have a period of oscillation of 1.00 s? The moment of inertia of a hoop of mass M and radius R about a point on its rim is 2MR2.

A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m undergoes simple harmonic oscillations with an amplitude of 7.50 cm.What is the total mechanical energy of the system?

A simple pendulum has a period T on Earth.If it were used on Planet X,where the acceleration due to gravity is 3 times what it is on Earth,what would its period be?

In 1851 Jean Bernard Leon Foucault demonstrated the rotation of the earth using a pendulum 11.0 m long,which was set up in the Paris Observatory.How long would it have taken for Foucault's pendulum to make one complete swing back to its starting point if g = 9.81 m/s2 at the observatory?

If both the mass of a simple pendulum and its length are doubled,the period will

How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 N/m so that it will oscillate at 1.00 Hz?

A thin hoop is supported in a vertical plane by a nail.What should the radius of the hoop be in order for it to have a period of oscillation of 1.00 s? The moment of inertia of a hoop of mass M and radius R about a point on its rim is 2MR².

In 1851 Jean Bernard Leon Foucault demonstrated the rotation of the earth using a pendulum 11.0 m long,which was set up in the Paris Observatory.How long would it have taken for Foucault's pendulum to make one complete swing back to its starting point if g = 9.81 m/s² at the observatory?