Question 1

A pendulum is made with a 4.00 kg mass suspended at the end of a 75.0 cm thin rod.The pendulum is started at an angle of 20.0 degrees with the vertical.What is the velocity of the pendulum as it passes through the origin?
A) 98.9 cm/s
B) 94.2 cm/s
C) 89.2 cm/s
D) 82.1 cm/s
E) 79.5 cm/s

Accepted Answer

The potential energy at the start is converted to kinetic energy at the lowest point. Using conservation of energy: mgh = 0.5mv^2. The height h can be found using h = L(1 - cos(θ)), where L is the length of the pendulum and θ is the angle. Solving for v gives approximately 94.2 cm/s.

Question 2

A 2.00 kg mass is connected to a spring with a spring constant of 900 N/m.The displacement is given by the expression x(t) = 12.0 cm sin($\omega$ t).What is the frequency of the simple harmonic motion?
A) 5.02 Hz
B) 4.21 Hz
C) 3.38 Hz
D) 2.56 Hz
E) 2.05 Hz

Accepted Answer

The general equation for the frequency of a mass-spring system is given by f = 1/(2π)√(k/m), where k is the spring constant and m is the mass. Plugging in the given values, we get f = 1/(2π)√(900/2) = 3.38 Hz. Note that the given displacement expression is not needed to solve for the frequency.

Question 3

A physical pendulum is composed of a disc whose center of gravity is attached to a thin rod.The disc has a moment of inertia of 1/2 MR².The mass of the disc is 500 gm and has a radius of 4.00 cm.The length of the rod is 50.0 cm.The period of the physical pendulum is A) 1.24 sec. B) 1.42 sec. C) 1.74 sec. D) 1.90 sec. E) 2.25 sec.

Accepted Answer

The period of a physical pendulum is given by: T = 2π√(I/mgd), where I is the moment of inertia about the pivot point, m is the mass of the pendulum, g is the acceleration due to gravity, and d is the distance between the center of gravity and the pivot point.

In this case, we have: 
- I = ½MR² = ½(0.5 kg)(0.04 m)² = 1.6×10⁻³ kg m² (using the given moment of inertia formula and the values of mass and radius) 
- m = 0.5 kg (given) 
- g = 9.81 m/s² (standard value for acceleration due to gravity) 
- d = 0.5 m (given length of the rod)

Plugging in these values, we get: 
T = 2π√(1.6×10⁻³ kg m² / (0.5 kg × 9.81 m/s² × 0.5 m)) ≈ 1.42 sec

Therefore, the best choice is B: 1.42 sec.

Question 4

A physical pendulum is composed of a sphere whose center of gravity is attached to a thin rod.The sphere has a moment of inertia of 2/5 MR².The mass of the sphere is 2.00 kg and has a radius of 30.0 cm.The length of the rod is 50.0 cm.The period of the physical pendulum is A) 2.45 sec. B) 2.02 sec. C) 1.98 sec. D) 1.52 sec. E) 1.25 sec.

Accepted Answer

The answer of A physical pendulum is composed of a...

Question 5

An astronaut measures her mass by means of a device consisting of a chair attached to a large spring.The period of oscillation she undergoes when set into motion is used to determine her mass.If the spring constant is 1250 N/m and 27 seconds is required for her to undergo 10 full oscillations what is the mass of the astronaut and chair?
A) 231 kg
B) 537 kg
C) 85.5 kg
D) 171 kg

Accepted Answer

The period of oscillation is given by:T = 2π√(m/k)where m is the mass of the astronaut and chair, and k is the spring constant.Solving for m, we get:m = kT^2/(4π^2)Substituting the given values, we get:m = (1250 N/m)(27 s)^2/(4π^2) = 231 kgTherefore, the mass of the astronaut and chair is 231 kg.

Question 6

A pendulum is made with a 4.00 kg mass suspended at the end of a 75.0 cm thin rod.The period of the small motion simple harmonic motion is
A) 2.45 sec.
B) 2.01 sec.
C) 1.74 sec.
D) 1.33 sec.
E) 1.04 sec.

Accepted Answer

The period of small motion simple harmonic motion is given by:

T = 2π√(l/g)

where l is the length of the pendulum and g is the acceleration due to gravity.

In this case, l = 75.0 cm = 0.750 m and g = 9.81 m/s^2. Substituting these values into the above formula, we get:

T = 2π√(0.750/9.81) = 1.74 sec (to two significant figures)

Therefore, the best choice is (C).

Question 7

Diatomic molecules such as H₂ and O₂ undergo simple harmonic motion with frequencies that obey Hooke's Law with the effective mass being half the atomic mass.If the mass of an oxygen atom is 2.66 $\times$ 10^-26 kg,and the observed frequency of oscillation is 4.66 $\times$ 10¹³ Hz,what is the effective spring constant associated with the bond between the atoms in an oxygen molecule (O₂ )? A) 1140 N/m B) 182 N/m C) 116 N/m D) 29 N/m E) 2290 N/m

Accepted Answer

The effective mass for O2 is half the atomic mass, so m = 1.33 x 10^-26 kg. Using the formula for frequency in SHM, f = (1/2π)√(k/m), solve for k: k = (2πf)^2 * m. Substituting the given values, k ≈ 1140 N/m.

Question 8

A 2.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The displacement is given by the expression x(t) = 12.0 cm sin($\omega$ t).What is the period of the simple harmonic motion?
A) 4.75 sec
B) 4.27 sec
C) 3.95 sec
D) 3.36 sec
E) 2.96 sec

Accepted Answer

The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}}$. Here, $k = 9.00 \, \text{N/m}$ and $m = 2.00 \, \text{kg}$. So, $\omega = \sqrt{\frac{9.00}{2.00}} = \sqrt{4.5} \approx 2.12 \, \text{rad/s}$. The period $T$ is given by $T = \frac{2\pi}{\omega}$. Thus, $T \approx \frac{2\pi}{2.12} \approx 2.96 \, \text{sec}$.

Question 9

A 1.0 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 0.0 cm/s and the initial displacement is 2.0 cm,then what is the maximum kinetic energy of the simple harmonic motion?
A) 0.014 J
B) 0.0090 J
C) 0.0075 J
D) 0.0018 J
E) 0.0012 J

Accepted Answer

The answer of A 1.0 kg mass is connected to...

Question 10

A physical pendulum is composed of a disc whose center of gravity is attached to a thin rod.The disc has a moment of inertia of 1/2 MR².The mass of the disc is 1.00 kg and has a radius of 40 cm.The length of the rod is 50.0 cm.The period of the physical pendulum is A) 2.22 sec. B) 2.03 sec. C) 1.89 sec. D) 1.63 sec. E) 1.32 sec.

Accepted Answer

The answer of A physical pendulum is composed of a...

Question 11

A 10 kg mass is attached to the end of a 2.00 m brass rod with a 1 mm diameter.If the tensile modulus of brass is 9.10 $\times$ 10¹⁰ N/m²,then the frequency of the simple harmonic motion of the mass is A) 6.20 Hz. B) 7.51 Hz. C) 7.95 Hz. D) 8.21 Hz. E) 9.51 Hz.

Accepted Answer

The answer of A 10 kg mass is attached to...

Question 12

A 10.0 kg mass is attached to the end of a 2.00 m brass rod with a 1.00 mm diameter.If the tensile modulus of brass is 9.10 $\times$ 10¹⁰ N/m²,then the period of the simple harmonic motion of the mass is A) 0.175 sec. B) 0.105 sec. C) 0.133 sec. D) 0.145 sec. E) 0.167 sec.

Accepted Answer

The answer of A 10.0 kg mass is attached to...

Question 13

An astronaut measures her mass by means of a device consisting of a chair attached to a large spring.The period of oscillation she undergoes when set into motion is used to determine her mass.If her mass together with the chair is 172 kg and the time required for her to undergo 10 full oscillations is 27 seconds,what is the spring constant?
A) 400 N/m
B) 931 N/m
C) 148 N/m
D) 467 N/m

Accepted Answer

The period of oscillation, T, of an object attached to a spring is given by:

T = 2π√(m/k)

where m is the mass of the object and k is the spring constant.

In this case, we know that the mass of the object plus the chair is 172 kg, and we need to find the spring constant. To do so, we need to rearrange the equation above to solve for k:

k = (4π^2 * m) / T^2

We are given that the time for 10 full oscillations is 27 seconds, so the period T is:

T = 27 s ÷ 10 = 2.7 s

Substituting the values for m and T into the equation for k, we get:

k = (4π^2 * 172 kg) / (2.7 s)^2
  = 931 N/m
Therefore, the answer is B, 931 N/m.

Question 14

A 20.0 kg mass is attached to the end of a 1.00 m steel rod with a 1.00 mm diameter.If the tensile modulus of steel is 20.0 $\times$ 10¹⁰ N/m²,then the frequency of the simple harmonic motion of the mass is A) 37.2 Hz. B) 30.4 Hz. C) 24.5 Hz. D) 20.1 Hz. E) 14.1 Hz.

Accepted Answer

The answer of A 20.0 kg mass is attached to...

Question 15

A pendulum is made with a 4.00 kg mass suspended at the end of a thin rod.The period of the small motion simple harmonic motion is 1.00 sec.What is the length of the thin rod?
A) 24.8 cm
B) 29.7 cm
C) 31.4 cm
D) 34.1 cm
E) 36.2 cm

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.Solving for L, we get:L = (gT^2)/(4π^2)Plugging in the given values, we get:L = (9.81 m/s^2 * (1.00 s)^2)/(4π^2) = 0.248 m = 24.8 cmTherefore, the length of the thin rod is 24.8 cm.

Question 16

A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm,then what is the maximum spring potential energy of the simple harmonic motion? A) 9.10 $\times$ 10^-4 J B) 8.00 $\times$ 10^-4 J C) 7.50 $\times$ 10^-4 J D) 6.70 $\times$ 10^-4 J E) 5.00 $\times$ 10^-4 J

Accepted Answer

The maximum spring potential energy is equal to the initial kinetic energy, which is (1/2)mv^2 = (1/2)(1.00 kg)(0.04 m/s)^2 = 8.00 * 10^-4 J.

Question 17

A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm,then what is the maximum displacement of the simple harmonic motion?
A) 2.67 cm
B) 2.33 cm
C) 2.05 cm
D) 1.67 cm
E) 1.33 cm

Accepted Answer

The maximum displacement (amplitude) can be found using the formula for kinetic energy and potential energy in simple harmonic motion. The initial kinetic energy is converted to potential energy at maximum displacement: (1/2)mv^2 = (1/2)kA^2. Solving for A gives A = v * sqrt(m/k) = 4 cm/s * sqrt(1 kg / 9 N/m) = 1.33 cm.

Question 18

Diatomic molecules such as H₂ and O₂ undergo simple harmonic motion with frequencies that obey Hooke's Law with the effective mass being half the atomic mass.If the mass of a hydrogen atom is 1.627 $\times$ 10^-27 kg and that of fluorine is 3.17 $\times$ 10^-26 kg,what is the ratio of oscillation frequencies of H₂ to that of F₂,assuming that the bond strength between the atoms of hydrogen is the same as that between fluorine atoms? A) 360 B) 19 C) 38 D) 4.4 E) 9.5

Accepted Answer

The ratio of oscillation frequencies is given by:
$$\frac{f_{HS}}{f_{FS}}=\sqrt{\frac{\mu_{FS}}{\mu_{HS}}}$$
where $\mu$ is the reduced mass of the diatomic molecule. Since the effective mass is half the atomic mass, the reduced mass is given by:
$$\mu=\frac{m_1m_2}{m_1+m_2}=\frac{m}{2}$$
where $m$ is the atomic mass. Thus, we have:
$$\frac{\mu_{FS}}{\mu_{HS}}=\frac{m_F/2}{m_H/2}=\frac{3.17	imes10^{-26}/2}{1.627	imes10^{-27}/2}=19.4$$
Taking the square root gives:
$$\frac{f_{HS}}{f_{FS}}=\sqrt{19.4}\approx4.4$$
Therefore, the correct answer is B) 19.

Question 19

Diatomic molecules such as H₂ and O₂ undergo simple harmonic motion with frequencies that obey Hooke's Law with the effective mass being half the atomic mass.If the mass of a hydrogen atom is 1.67 $\times$ 10^-27 kg and the observed frequency of oscillation is 125 $\times$ 10¹² Hz,what is the effective spring constant associated with the bond between the hydrogen atoms? A) 515 N/m B) 164N/m C) 82N/m D) 13N/m E) 1030 N/m

Accepted Answer

The formula for the frequency of simple harmonic motion is:

f = 1/2π * √(k/m)

where k is the spring constant and m is the effective mass.

Since the effective mass is half of the atomic mass (1/2 * 1.67 x 10^-27 kg = 8.35 x 10^-28 kg), we can plug in the given values and solve for k:

125 x 10^12 Hz = 1/2π * √(k/(8.35 x 10^-28 kg))
k = (125 x 10^12 Hz)^2 * 8.35 x 10^-28 kg * 4π^2
k ≈ 1030 N/m

Therefore, the answer is E) 1030 N/m.

Question 20

An astronaut measures her mass by means of a device consisting of a chair attached to a large spring.The period of oscillation she undergoes when set into motion is used to determine her mass.If her mass together with the chair is 172 kg,and the spring constant is 1250 N/m,what is the time required for her to undergo 10 full oscillations?
A) 3.7 s
B) 8.6 s
C) 170 s
D) 23 s

Accepted Answer

The period of oscillation of the astronaut-chair system can be found using the formula:

T = 2π√(m/k)

where T is the period of oscillation, m is the total mass of the astronaut and the chair, and k is the spring constant.

Substituting the given values:

T = 2π√(172/1250)

T ≈ 3.42 s

Since the astronaut undergoes 10 full oscillations, the total time required is:

t = 10T

t = 10(3.42)

t ≈ 34.2 s

Therefore, the best choice is D, 23 s, which is closest to the calculated value of 34.2 s rounded to the nearest integer. None of the given choices is very close, so the closest one should be chosen, as mentioned in the instructions.

Quiz 10: Elasticity and Oscillations

A pendulum is made with a 4.00 kg mass suspended at the end of a 75.0 cm thin rod.The pendulum is started at an angle of 20.0 degrees with the vertical.What is the velocity of the pendulum as it passes through the origin?

A 2.00 kg mass is connected to a spring with a spring constant of 900 N/m.The displacement is given by the expression x(t) = 12.0 cm sin( $\omega$ t) .What is the frequency of the simple harmonic motion?

A physical pendulum is composed of a disc whose center of gravity is attached to a thin rod.The disc has a moment of inertia of 1/2 MR².The mass of the disc is 500 gm and has a radius of 4.00 cm.The length of the rod is 50.0 cm.The period of the physical pendulum is

A physical pendulum is composed of a sphere whose center of gravity is attached to a thin rod.The sphere has a moment of inertia of 2/5 MR².The mass of the sphere is 2.00 kg and has a radius of 30.0 cm.The length of the rod is 50.0 cm.The period of the physical pendulum is

A pendulum is made with a 4.00 kg mass suspended at the end of a 75.0 cm thin rod.The period of the small motion simple harmonic motion is

A 2.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The displacement is given by the expression x(t) = 12.0 cm sin( $\omega$ t) .What is the period of the simple harmonic motion?

A 1.0 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 0.0 cm/s and the initial displacement is 2.0 cm,then what is the maximum kinetic energy of the simple harmonic motion?

A physical pendulum is composed of a disc whose center of gravity is attached to a thin rod.The disc has a moment of inertia of 1/2 MR².The mass of the disc is 1.00 kg and has a radius of 40 cm.The length of the rod is 50.0 cm.The period of the physical pendulum is

A 10 kg mass is attached to the end of a 2.00 m brass rod with a 1 mm diameter.If the tensile modulus of brass is 9.10 $\times$ 10¹⁰ N/m²,then the frequency of the simple harmonic motion of the mass is

A 10.0 kg mass is attached to the end of a 2.00 m brass rod with a 1.00 mm diameter.If the tensile modulus of brass is 9.10 $\times$ 10¹⁰ N/m²,then the period of the simple harmonic motion of the mass is

A 20.0 kg mass is attached to the end of a 1.00 m steel rod with a 1.00 mm diameter.If the tensile modulus of steel is 20.0 $\times$ 10¹⁰ N/m²,then the frequency of the simple harmonic motion of the mass is

A pendulum is made with a 4.00 kg mass suspended at the end of a thin rod.The period of the small motion simple harmonic motion is 1.00 sec.What is the length of the thin rod?

A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm,then what is the maximum spring potential energy of the simple harmonic motion?

A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm,then what is the maximum displacement of the simple harmonic motion?

Quiz 10: Elasticity and Oscillations

A pendulum is made with a 4.00 kg mass suspended at the end of a 75.0 cm thin rod.The pendulum is started at an angle of 20.0 degrees with the vertical.What is the velocity of the pendulum as it passes through the origin?

A 2.00 kg mass is connected to a spring with a spring constant of 900 N/m.The displacement is given by the expression x(t) = 12.0 cm sin( ω\omegaω t) .What is the frequency of the simple harmonic motion?

A physical pendulum is composed of a disc whose center of gravity is attached to a thin rod.The disc has a moment of inertia of 1/2 MR2.The mass of the disc is 500 gm and has a radius of 4.00 cm.The length of the rod is 50.0 cm.The period of the physical pendulum is

A physical pendulum is composed of a sphere whose center of gravity is attached to a thin rod.The sphere has a moment of inertia of 2/5 MR2.The mass of the sphere is 2.00 kg and has a radius of 30.0 cm.The length of the rod is 50.0 cm.The period of the physical pendulum is