Question 1

A copper ball with a diameter of 5.00 cm is subjected to a pressure of 2.50 $\times$ 10⁶ Pa.The bulk modulus for copper is 14.00 $\times$ 10¹⁰ N/m².What is the bulk strain? A) 1.79 $\times$ 10^-5 B) 2.09 $\times$ 10^-5 C) 2.55 $\times$ 10^-5 D) 2.94 $\times$ 10^-5 E) 3.21 $\times$ 10^-5

Accepted Answer

Bulk strain is calculated using the formula: Bulk strain = Pressure / Bulk modulus. Substituting the given values: (2.50 * 10^6 Pa) / (14.00 * 10^10 N/m^2) = 1.79 * 10^-5.

Question 2

A 4.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 12.0 cm/s and the initial displacement is 4.00 cm,then what is the maximum velocity of the simple harmonic motion?
A) 20.3 cm/s
B) 19.8 cm/s
C) 15.5 cm/s
D) 13.4 cm/s
E) 11.5 cm/s

Accepted Answer

The answer of A 4.00 kg mass is connected to...

Question 3

A 2.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin($\omega$ t).What is the maximum acceleration of the simple harmonic motion? A) 17.5 cm/s² B) 21.2 cm/s² C) 25.2 cm/s² D) 30.5 cm/s² E) 36.7 cm/s²

Accepted Answer

The maximum acceleration of a mass on a spring is given by the formula:

a_max = w^2 * A

where w is the angular frequency (2*pi*f) and A is the amplitude of the motion.

In this case, we can find the angular frequency by noting that the velocity function is given by:

v(t) = A*w*cos(w*t)

Comparing this to the given velocity function, we see that:

A*w = 10.0 cm/s

and

w = 2*pi*f

So,

f = w/(2*pi) = (10.0 cm/s)/(2*pi*(0.01 m)) = 1.59 Hz

Now we can find the amplitude of the motion by noting that the maximum speed is equal to the amplitude:

v_max = A*w

A = v_max/w = (10.0 cm/s)/((2*pi)*(1.59 Hz)) = 0.994 cm

Finally, we can find the maximum acceleration:

a_max = w^2*A = ((2*pi*f)^2)*(0.994 cm) = 21.2 cm/s^2

Question 4

A steel ball with a diameter of 10 cm is subjected to a pressure that causes the volume of the ball to shrink by 0.01%.The bulk modulus for steel is 16.0 $\times$ 10¹⁰ N/m².What is the pressure? A) 2.4 $\times$ 10⁷ N/m² B) 2.1 $\times$ 10⁷ N/m² C) 1.9 $\times$ 10⁷ N/m² D) 1.6 $\times$ 10⁷ N/m² E) 1.2 $\times$ 10⁷ N/m²

Accepted Answer

The answer of A steel ball with a diameter of...

Question 5

An aluminum ball is subjected to a pressure of 7.00 $\times$ 10⁶ Pa.The bulk strain is 1.0 $\times$ 10^-4.What is the bulk modulus? A) 5.8 $\times$ 10¹⁰ N/m² B) 6.1 $\times$ 10¹⁰ N/m² C) 6.3 $\times$ 10¹⁰ N/m² D) 7.0 $\times$ 10¹⁰ N/m² E) 8.1 $\times$ 10¹⁰ N/m²

Accepted Answer

The answer of An aluminum ball is subjected to a...

Question 6

A mass is suspended vertically from a spring so it is at rest at the equilibrium position.The mass is pulled straight down to an extension x and released so that it oscillates about the equilibrium position.The acceleration is greatest in magnitude and directed downward when the mass is
A) at its maximum upward travel.
B) at the equilibrium point.
C) at its maximum lower travel.
D) somewhere between the equilibrium point and maximum extension.

Accepted Answer

The answer of A mass is suspended vertically from a...

Question 7

A 4.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 8.00 cm/s sin($\omega$ t)+ 10.0 cm/s cos($\omega$ t).What is the maximum velocity of the simple harmonic motion?
A) 41.0 cm/s
B) 32.8 cm/s
C) 24.5 cm/s
D) 20.5 cm/s
E) 12.8 cm/s

Accepted Answer

The answer of A 4.00 kg mass is connected to...

Question 8

A 9.00 kg mass is connected to a spring with a spring constant of 4.00 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin($\omega$ t)+ 8.00 cm/s cos($\omega$ t).What is the maximum displacement of the simple harmonic motion?
A) 25.1 cm
B) 30.5 cm
C) 35.3 cm
D) 40.4 cm
E) 45.5 cm

Accepted Answer

The maximum displacement (amplitude) can be found using the formula $ A = \sqrt{v_0^2 + u_0^2} / \omega $, where $ v_0 = 10.0 $ cm/s, $ u_0 = 8.00 $ cm/s, and $ \omega = \sqrt{k/m} = \sqrt{4/9} $ rad/s. Calculating gives $ A = \sqrt{10^2 + 8^2} / \sqrt{4/9} = 25.1 $ cm.

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 4.0 cm/s and the initial displacement is 2.0 cm,then what is the maximum spring potential energy of the simple harmonic motion?
A) 0.0026 J
B) 0.0059 J
C) 0.0035 J
D) 0.0067 J
E) 0.0060 J

Accepted Answer

The maximum spring potential energy of the simple harmonic motion is equal to the initial kinetic energy of the mass. The initial kinetic energy is given by:$$KE = \frac{1}{2}mv^2 = \frac{1}{2}(1.0 	ext{ kg})(0.04 	ext{ m/s})^2 = 0.0008 	ext{ J}$$Therefore, the maximum spring potential energy is also 0.0008 J.

Question 10

A mass is suspended vertically from a spring so it is at rest at the equilibrium position.The mass is pulled straight down to an extension x and released so that it oscillates about the equilibrium position.The acceleration of the mass is zero when the mass is
A) at its maximum upward travel.
B) at the equilibrium point.
C) at its maximum lower travel.
D) somewhere between the equilibrium point and maximum extension.

Accepted Answer

At the equilibrium point, the spring force and the weight of the mass balance each other out, resulting in a net force of zero. This means that the acceleration of the mass is also zero at this point. The acceleration is maximum at the extreme points of the motion (at maximum upward and maximum lower travel), where the net force is the greatest. Therefore, the correct answer is B.

Question 11

A 2.00 kg mass is connected to a spring with a spring constant of 6.00 N/m.The displacement is given by the expression x(t) = 12.0 cm sin($\omega$ t).What is the maximum velocity the simple harmonic motion?
A) 34.8 cm/s
B) 30.7 cm/s
C) 25.6 cm/s
D) 20.8 cm/s
E) 17.5 cm/s

Accepted Answer

The answer of A 2.00 kg mass is connected to...

Question 12

Seawater has a density of 1,025 kg/m³ at the surface.The bulk modulus of seawater is 0.2100 $\times$ 10¹⁰ N/m².What is the density of seawater at a depth of 10.00 km? A) 1074 kg/m³ B) 1230 kg/m³ C) 1574 kg/m³ D) 1764 kg/m³ E) 1973 kg/m³

Accepted Answer

The change in density can be calculated using the formula for bulk modulus: B = -V * (dP/dV). At 10 km depth, pressure increases by approximately 1.01 * 10^8 N/m^2. Using the bulk modulus, the density increases to approximately 1074 kg/m^3.

Question 13

An equation that describes the displacement in Simple Harmonic Motion is: x(t) = 1.20 m sin(2.40 rad/s t).What is the maximum velocity of the SHM?
A) 5.32 m/s
B) 4.82 m/s
C) 3.68 m/s
D) 2.88 m/s
E) 2.03 m/s

Accepted Answer

The answer of An equation that describes the displacement in...

Question 14

A 1.0 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 2.00 cm,then what is the maximum kinetic energy of the simple harmonic motion?
A) 0.0023 J
B) 0.0078 J
C) 0.0026 J
D) 0.0020 J
E) 0.0012 J

Accepted Answer

The maximum kinetic energy in simple harmonic motion is given by the total mechanical energy, which is the sum of initial kinetic and potential energies. KE_initial = 0.5 * m * v^2 = 0.5 * 1 * (0.04)^2 = 0.0008 J. PE_initial = 0.5 * k * x^2 = 0.5 * 9 * (0.02)^2 = 0.0018 J. Total energy = 0.0008 J + 0.0018 J = 0.0026 J.

Question 15

A mass is suspended vertically from a spring so it is at rest at the equilibrium position.The mass is pulled straight down to an extension x and released so that it oscillates about the equilibrium position.The speed of the mass is greatest when the mass is
A) at its maximum upward travel.
B) at the equilibrium point.
C) at its maximum lower travel.
D) somewhere between the equilibrium point and maximum extension.

Accepted Answer

The speed of the mass is greatest at the equilibrium point because all the potential energy stored in the spring is converted to kinetic energy at this point, resulting in maximum speed.

Question 16

A 3.0 kg mass is connected to a spring with a spring constant of 6.0 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin($\omega$ t).What is the maximum velocity of the simple harmonic motion?
A) 10 cm/s
B) 14 cm/s
C) 17 cm/s
D) 20 cm/s
E) 25 cm/s

Accepted Answer

The maximum velocity in simple harmonic motion is given by the amplitude of the velocity function. Here, the amplitude is 10 cm/s, which is the coefficient of the sine function in v(t).

Question 17

A copper ball with a diameter of 5.00 cm is subjected to a pressure of 2.50 $\times$ 10⁶ Pa.The bulk modulus for copper is 14.00 $\times$ 10¹⁰ N/m².What is the change in the volume? A) 1.01 $\times$ 10^-3 cm³ B) 1.17 $\times$ 10^-3 cm³ C) 1.20 $\times$ 10^-3 cm³ D) 1.58 $\times$ 10^-3 cm³ E) 1.70 $\times$ 10^-3 cm³

Accepted Answer

The change in volume can be calculated using the formula ΔV = - (ΔP * V) / B, where ΔP is the pressure change, V is the initial volume, and B is the bulk modulus. The initial volume V of the sphere is (4/3)πr³, where r is the radius (2.5 cm). Convert the volume to m³, calculate ΔV, and convert back to cm³. The correct answer is approximately 1.17 * 10⁻³ cm³.

Question 18

A 4.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 8.00 cm/s sin($\omega$t)+ 10.0 cm/s cos($\omega$ t).What is the maximum acceleration of the simple harmonic motion? A) 10.0 cm/s² B) 16.2 cm/s² C) 19.2 cm/s² D) 20.2 cm/s² E) 24.5 cm/s²

Accepted Answer

The maximum acceleration in simple harmonic motion is given by $ a_{max} = \omega^2 A $, where $ A $ is the amplitude of the velocity. The angular frequency $ \omega $ is calculated using $ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{9.00}{4.00}} = 1.5 \, \text{rad/s} $. The amplitude $ A $ of the velocity is $ \sqrt{(8.00)^2 + (10.0)^2} = \sqrt{164} \approx 12.8 \, \text{cm/s} $. Thus, $ a_{max} = (1.5)^2 \times 12.8 \approx 19.2 \, \text{cm/s}^2 $.

Question 19

A 2.0 kg mass is connected to a spring with a spring constant of 9.0 N/m.The displacement is given by the expression x(t) = 12.0 cm sin($\omega$ t).What is the maximum displacement of the simple harmonic motion?
A) 8.0 cm
B) 12 cm
C) 20 cm
D) 24 cm
E) 30 cm

Accepted Answer

The maximum displacement in simple harmonic motion is the amplitude of the motion, which is given directly in the expression x(t) = 12.0 cm sin(ωt). Therefore, the maximum displacement is 12 cm.

Question 20

A mass is suspended vertically from a spring so it is at rest at the equilibrium position.The mass is pulled straight down to an extension x and released so that it oscillates about the equilibrium position.The acceleration is greatest in magnitude and directed upward when the mass is
A) at its maximum upward travel.
B) at the equilibrium point.
C) at its maximum lower travel.
D) somewhere between the equilibrium point and maximum extension.

Accepted Answer

The acceleration is greatest in magnitude when the spring force is greatest, which occurs at the maximum extension (maximum lower travel) due to Hooke's Law (F = -kx). At this point, the acceleration is directed upward as the spring force acts to restore the mass to equilibrium.

Quiz 10: Elasticity and Oscillations

A copper ball with a diameter of 5.00 cm is subjected to a pressure of 2.50 $\times$ 10⁶ Pa.The bulk modulus for copper is 14.00 $\times$ 10¹⁰ N/m².What is the bulk strain?

A 4.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 12.0 cm/s and the initial displacement is 4.00 cm,then what is the maximum velocity of the simple harmonic motion?

A 2.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin( $\omega$ t) .What is the maximum acceleration of the simple harmonic motion?

A steel ball with a diameter of 10 cm is subjected to a pressure that causes the volume of the ball to shrink by 0.01%.The bulk modulus for steel is 16.0 $\times$ 10¹⁰ N/m².What is the pressure?

An aluminum ball is subjected to a pressure of 7.00 $\times$ 10⁶ Pa.The bulk strain is 1.0 $\times$ 10^-4.What is the bulk modulus?

A mass is suspended vertically from a spring so it is at rest at the equilibrium position.The mass is pulled straight down to an extension x and released so that it oscillates about the equilibrium position.The acceleration is greatest in magnitude and directed downward when the mass is

A 4.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 8.00 cm/s sin( $\omega$ t) + 10.0 cm/s cos( $\omega$ t) .What is the maximum velocity of the simple harmonic motion?

A 9.00 kg mass is connected to a spring with a spring constant of 4.00 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin( $\omega$ t) + 8.00 cm/s cos( $\omega$ t) .What is the maximum displacement 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 4.0 cm/s and the initial displacement is 2.0 cm,then what is the maximum spring potential energy of the simple harmonic motion?

A mass is suspended vertically from a spring so it is at rest at the equilibrium position.The mass is pulled straight down to an extension x and released so that it oscillates about the equilibrium position.The acceleration of the mass is zero when the mass is

A 2.00 kg mass is connected to a spring with a spring constant of 6.00 N/m.The displacement is given by the expression x(t) = 12.0 cm sin( $\omega$ t) .What is the maximum velocity the simple harmonic motion?

Seawater has a density of 1,025 kg/m³ at the surface.The bulk modulus of seawater is 0.2100 $\times$ 10¹⁰ N/m².What is the density of seawater at a depth of 10.00 km?

An equation that describes the displacement in Simple Harmonic Motion is: x(t) = 1.20 m sin(2.40 rad/s t) .What is the maximum velocity of the SHM?

A 1.0 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 2.00 cm,then what is the maximum kinetic energy of the simple harmonic motion?

A mass is suspended vertically from a spring so it is at rest at the equilibrium position.The mass is pulled straight down to an extension x and released so that it oscillates about the equilibrium position.The speed of the mass is greatest when the mass is

A 3.0 kg mass is connected to a spring with a spring constant of 6.0 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin( $\omega$ t) .What is the maximum velocity of the simple harmonic motion?

A copper ball with a diameter of 5.00 cm is subjected to a pressure of 2.50 $\times$ 10⁶ Pa.The bulk modulus for copper is 14.00 $\times$ 10¹⁰ N/m².What is the change in the volume?

A 4.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 8.00 cm/s sin( $\omega$ t) + 10.0 cm/s cos( $\omega$ t) .What is the maximum acceleration of the simple harmonic motion?

A 2.0 kg mass is connected to a spring with a spring constant of 9.0 N/m.The displacement is given by the expression x(t) = 12.0 cm sin( $\omega$ t) .What is the maximum displacement of the simple harmonic motion?

A mass is suspended vertically from a spring so it is at rest at the equilibrium position.The mass is pulled straight down to an extension x and released so that it oscillates about the equilibrium position.The acceleration is greatest in magnitude and directed upward when the mass is

Quiz 10: Elasticity and Oscillations

A copper ball with a diameter of 5.00 cm is subjected to a pressure of 2.50 ×\times× 106 Pa.The bulk modulus for copper is 14.00 ×\times× 1010 N/m2.What is the bulk strain?

A 4.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 12.0 cm/s and the initial displacement is 4.00 cm,then what is the maximum velocity of the simple harmonic motion?

A 2.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin( ω\omegaω t) .What is the maximum acceleration of the simple harmonic motion?

A steel ball with a diameter of 10 cm is subjected to a pressure that causes the volume of the ball to shrink by 0.01%.The bulk modulus for steel is 16.0 ×\times× 1010 N/m2.What is the pressure?

An aluminum ball is subjected to a pressure of 7.00 ×\times× 106 Pa.The bulk strain is 1.0 ×\times× 10-4.What is the bulk modulus?