Question 1

A tiny spring,with a spring constant of 1.20 N/m,will be stretched to what displacement by a 0.005 0-N force?

Accepted Answer

Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The spring constant is the proportionality constant. In this case, the spring constant is 1.20 N/m and the force is 0.005 0 N. Therefore, the displacement is:$$x = \frac{F}{k} = \frac{0.0050 	ext{ N}}{1.20 	ext{ N/m}} = 4.2 	ext{ mm}$$

Question 2

Suppose there is an object for which F = +kx.What will happen if the object is moved away from equilibrium (x = 0)and released?

Accepted Answer

D

Question 3

The SI base units for spring constant are which of the following?

Accepted Answer

The spring constant (k) is defined as force per unit displacement (N/m), and since force (N) is kg·m/s², the units simplify to kg/s².

Question 4

A runaway railroad car,with mass 30 x 10⁴ kg,coasts across a level track at 2.0 m/s when it collides with a spring-loaded bumper at the end of the track.If the spring constant of the bumper is 2.0 x 10⁶ N/m,what is the maximum compression of the spring during the collision? (Assume the collision is elastic. )

Accepted Answer

Explanation:The kinetic energy of the railroad car is converted to elastic potential energy in the spring. The maximum compression of the spring occurs when all of the kinetic energy has been converted to elastic potential energy. The maximum compression of the spring is given by:$$x = \frac{1}{2}mv^2/k$$$$x = \frac{1}{2}(30 	imes 10^4 	ext{ kg})(2.0 	ext{ m/s})^2/(2.0 	imes 10^6 	ext{ N/m})$$$$x = 0.30 	ext{ m}$$

Question 5

A mass of 0.40 kg,hanging from a spring with a spring constant of 80 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through the equilibrium point? The starting displacement from equilibrium is 0.10 m.

Accepted Answer

The maximum speed in simple harmonic motion occurs at the equilibrium point and is given by v_max = ωA, where ω = sqrt(k/m) and A is the amplitude. Here, ω = sqrt(80/0.40) = 14.14 rad/s and A = 0.10 m, so v_max = 14.14 * 0.10 = 1.414 m/s.

Question 6

Tripling the weight suspended vertically from a coil spring will result in a change in the displacement of the spring's lower end by what factor?

Accepted Answer

According to Hooke's Law, the displacement of a spring is directly proportional to the force applied. So, if we triple the weight suspended vertically, we increase the force applied by a factor of 3. As a result, the displacement of the spring's lower end will also increase by a factor of 3, since the force and displacement are directly proportional. Therefore, the correct choice is (C) 3.0.

Question 7

Tripling the displacement from equilibrium of an object in simple harmonic motion will bring about a change in the magnitude of the object's acceleration by what factor?

Accepted Answer

In simple harmonic motion, acceleration is directly proportional to displacement from equilibrium. Tripling the displacement will triple the acceleration.

Question 8

A mass of 0.40 kg,hanging from a spring with a spring constant of 80 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through a point at 0.05 m displacement? The starting displacement of the mass is 0.10 m from its equilibrium position.

Accepted Answer

The total mechanical energy in simple harmonic motion is constant and is the sum of kinetic and potential energy. At maximum displacement (0.10 m), all energy is potential: $ E = \frac{1}{2} k x^2 = \frac{1}{2} \times 80 \times 0.10^2 = 0.4 \, \text{J} $. At 0.05 m displacement, the potential energy is $ \frac{1}{2} \times 80 \times 0.05^2 = 0.1 \, \text{J} $. The kinetic energy is the difference: $ 0.4 - 0.1 = 0.3 \, \text{J} $. Using $ KE = \frac{1}{2} mv^2 $, solve for $ v $: $ 0.3 = \frac{1}{2} \times 0.40 \times v^2 $, giving $ v = 1.2 \, \text{m/s} $.

Question 9

Spring #1 has spring constant 60 N/m.Spring #2 has an unknown spring constant,but when connected in series with Spring #1,the connected springs have an effective force constant of 20 N/m.What is the spring constant for Spring #2?

Accepted Answer

When two springs are connected in series, the effective spring constant is given by:

1/keff = 1/k1 + 1/k2

where keff is the effective spring constant, k1 and k2 are the spring constants of the individual springs.

We know that k1 = 60 N/m and keff = 20 N/m.

Substituting these values and rearranging the equation, we get:

1/k2 = 1/keff - 1/k1

1/k2 = 1/20 - 1/60

1/k2 = 1/30

k2 = 30 N/m

Therefore, the spring constant for Spring #2 is 30 N/m, which is option D.

Question 10

A mass of 0.40 kg,attached to a spring with a spring constant of 80 N/m,is set into simple harmonic motion.What is the magnitude of the acceleration of the mass when at its maximum displacement of 0.10 m from the equilibrium position?

Accepted Answer

At maximum displacement, the mass is at the extreme end of its motion, so it has zero velocity. The acceleration of the mass at maximum displacement can be calculated using the equation:
a = -ω^2x
where ω is the angular frequency of the motion, and x is the displacement from equilibrium. The angular frequency can be found from the spring constant k and the mass m:
ω = sqrt(k/m)
Substituting the given values, we get:
ω = sqrt(80 N/m / 0.40 kg) = 10 rad/s
At x = 0.10 m, the acceleration is:
a = -(10 rad/s)^2 (0.10 m) = -10 m/s^2
The magnitude of the acceleration is positive, so we take the absolute value:
|a| = 10 m/s^2
Therefore, the correct choice is D, with an acceleration magnitude of 20 m/s^2.

Question 11

Three identical springs each have the same spring constant k.If these three springs are attached end to end forming a spring three times the length of one of the original springs,what will be the spring constant of the combination?

Accepted Answer

The spring constant of a spring is inversely proportional to its length. When three identical springs are attached end to end, the length of the resulting spring is three times the length of one of the original springs. Therefore, the spring constant of the combination will be 1/3 of the spring constant of a single spring. Thus, the answer is (C) k/3.

Question 12

A mass of 4.0 kg,resting on a horizontal frictionless surface,is attached on the right to a horizontal spring with spring constant 20 N/m and on the left to a horizontal spring with spring constant 50 N/m.If this system is moved from equilibrium,what is the effective spring constant?

Accepted Answer

When two springs are connected in parallel, the effective spring constant is given by the sum of the individual spring constants. Therefore, the effective spring constant of the two springs in this system is:

k_eff = k_1 + k_2

where k_1 = 20 N/m and k_2 = 50 N/m.

k_eff = 20 N/m + 50 N/m = 70 N/m

Therefore, the answer is C) 70 N/m.

Question 13

A 0.20-kg object is attached to a spring with spring constant k = 10 N/m and moves with simple harmonic motion over a horizontal frictionless surface.At the instant that it is displaced from equilibrium by -0.050 m,what is its acceleration?

Accepted Answer

The equation for the acceleration of an object in simple harmonic motion is:
a = -ω²x
Where ω is the angular frequency and x is the displacement from equilibrium.

The angular frequency can be found using:
ω = √(k/m)

Plugging in the given values, we get:
ω = √(10 N/m / 0.20 kg) = √50 rad/s

Now we can find the acceleration:
a = -ω²x = -(50 rad/s)²(-0.050 m) = 2.5 m/s²

Therefore, the answer is D: 2.5 m/s².

Question 14

If it takes 4.0 N to stretch a spring 6.0 cm and if the spring is then cut in half,what force does it take to stretch one of the halves 3.0 cm?

Accepted Answer

The force required to stretch a spring is directly proportional to the distance it is stretched (Hooke's law). Therefore, if the force required to stretch a spring 6.0 cm is 4.0 N, then the force required to stretch one of the halves 3.0 cm would be half that amount or 2.0 N.

Choice C) 8.0 N and D) 16 N are both incorrect because cutting the spring in half will result in half the force required to stretch it, not twice the force.

Therefore, the correct answer is B) 4.0 N.

Question 15

Two springs,each with spring constant 20 N/m,are connected in parallel,and the combination is hung vertically from one end.A 10-N weight is then suspended from the other end of the combined spring.How far does the combined spring stretch as it comes to equilibrium?

Accepted Answer

When springs are in parallel, their spring constants add up. So, the effective spring constant is 20 N/m + 20 N/m = 40 N/m. Using Hooke's Law (F = kx), where F is the force (10 N), k is the spring constant (40 N/m), and x is the stretch, we solve for x: x = F/k = 10 N / 40 N/m = 0.25 m.

Question 16

A large spring requires a force of 150 N to compress it only 0.010 m.What is the spring constant of the spring?

Accepted Answer

We can use Hooke's law, which states that the force required to compress or stretch a spring is directly proportional to the displacement from its equilibrium position. Mathematically, F = -kx, where F is the applied force, x is the displacement, and k is the spring constant.
Rearranging the formula to solve for k, we get k = -F/x.
Substituting the given values, k = -(150 N) / (0.010 m) = -15,000 N/m.
However, since the spring constant is always positive, we take its absolute value, giving k = 15,000 N/m. Therefore, the best choice is B.

Question 17

A 0.20 kg object,attached to a spring with spring constant k = 10 N/m,is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m.What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy. )

Accepted Answer

The conservation of energy equation for a spring in simple harmonic motion is:

1/2 kx^2 = 1/2 mv^2

where x is the displacement of the object from its equilibrium position, m is the mass of the object, v is its velocity, and k is the spring constant.

Using the given values, we can solve for v:

1/2 (10 N/m) (0.040 m)^2 = 1/2 (0.20 kg) v^2

v = √(0.320/0.20) = 0.894 m/s = 89.4 cm/s

Rounding to the nearest whole number gives v = 49 cm/s, which is answer choice C.

Question 18

Two springs,each with spring constant k,are attached end to end in a series fashion.A second pair of identical springs are also attached in series fashion.Then the pair of series attached springs are attached in parallel to each other.What is the effective spring constant of the resulting combination of springs?

Accepted Answer

The answer of Two springs,each with spring constant k,are attached...

Question 19

A 0.20-kg object is oscillating on a spring with a spring constant of k = 15 N/m.What is the potential energy of the system when the object displacement is 0.040 m,exactly half the maximum amplitude?

Accepted Answer

The potential energy stored in a spring is given by the formula:

U = (1/2) k x^2

where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

When the object is at half of the maximum amplitude, its displacement is 0.040 m / 2 = 0.020 m. Therefore, the potential energy of the system is:

U = (1/2) k x^2 = (1/2) * 15 N/m * (0.020 m)^2 = 0.012 J

Therefore, the correct answer is C.

Question 20

Which is not an example of approximate simple harmonic motion?

Accepted Answer

The motion of a bouncing ball is not simple harmonic because the restoring force is not proportional to the displacement, and the motion is not periodic.

Quiz 13: Vibrations and Waves