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1. A mass is executing **simple harmonic motion** with amplitude A. What is the total distance traveled by the mass during one period of the oscillation? Does your answer depend on the time instant from which the period of the oscillation is measured? ANSWER

2. The following diagram depicts a series of straight waves traveling from shallow water into deeper water. Complete the diagram by drawing in the following:

(a) incident ray; (b) **reflected** ray; (c) **refracted** ray; (d) two refracted wave fronts; (e) two reflected wave fronts. ANSWER

3. The security alarm on a parked car goes off and produces a frequency of 735 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 78.4 Hz. At what speed are you driving? ANSWER

4. Transverse waves traveling across a rope have a **frequency** of 12.0 Hz and a **wavelength** of 2.40 m. What is the **velocity** of the **waves**? ANSWER

1.

During one complete oscillation, the mass travels a distance of 4A.

Where the cycle is deemed to start does not matter.

2.

3.

The change in frequency can be calculated using the general **Doppler** equation:

f'=f((V+-Vo) / (V-+Vs))

V is speed of sound (343m/s)

Vo is speed of observer

Vs is speed of source

f' is shifted frequency

f is original frequency

f'=f((V+Vo)/(V-Vs)) used for situations in which the source and the observer are coming closer together

f'=f((V-Vo)/(V+Vs)) used for situations that they are moving away from one another.

So for this question, when you drive toward this parked car

f'1=735*(343+Vo)/343

when you drive away

f'2=735*(343-Vo)/343

The difference in f' will be 735*2*Vo/343=78.4 , solve this one gives the speed of observer

Vo=18.29m/s=65.86km/hr

4.

The **universal wave equation** applies:

v = λf where v = the speed of propagation, λ = wavelength, and f = frequency.

v = λf = (12)(2.40) = 28.8 m/s

1. How can the speed of a spring undergoing simple harmonic motion be doubled for a particular amplitude?

2. A light spring is used to induce simple harmonic motion in a 5.7 kg block on a frictionless horizontal surface. When the block has a displacement of -0.70m, it has a velocity of -0.80 m/s and is accelerating at +2.7m/s^{2} . What is the amplitude of the motion?

3. The suspension of a 526 kg motorcycle is provided by springs with a combined force constant of 9067 N/m. A 110 kg rider will change the bike's oscillation period by what percentage, compared to the bike without a rider?

4. A mass attached to a spring starts from rest at x = 0.0410 m and time t = 0. If the mass oscillates with a period of 3.13 s, where is it at 3.21 s?

5. A vertical spring of force constant 162 N/m oscillates with a maximum speed of 0.372 m/s when attached to a 0.824 kg mass. Calculate

a. the period of the motion.

b. the maximum acceleration of the mass.

6. A mass attached to a spring exhibits simple harmonic motion with angular frequency, ω, and amplitude, A. What is the speed of the mass when its kinetic and potential energies are equal?

7. Why is the acceleration of vibrating body is zero at its equilibrium position?

8. A mass attached to a spring (k = 29.8 N/m) oscillates with an angular frequency of 2.81 rad/s. The spring's maximum distortion is 0.232 m. What is the potential energy stored in the system 1.42 s after the spring has extended to its maximum?

9. What length of **pendulum** near the Earth's surface will have a period of 1.0s?

1. The following shows a view from above an interference pattern. The two sources, S1 and S2, have the same frequency and are in phase. The point, P, is on the second nodal line. What is the wavelength of the sources?

2. Label compression, rarefaction, crest, trough, wavelength. Also describe the motion of points A and B.

3. The following diagrams are successive "snapshots" of two pulses travelling in opposite directions along a coil. Draw three diagrams to show the shape of the coil at t = 0.03 s, t = 0.06, and 0.09 s.

t = 0 s

t = 0.12 s

4. The following 2 pulses on a coil are pictured approaching a second coil. Draw a diagram to demonstrate the behavior of the pulses if the second coil is

a. faster than the first coil

b. slower than the first coil.

5. When combined with the pulse on the left, which of the pulses below will result in the following situation:

6. A certain vibration has a wavelength of 58.6m and travels at a speed of 1.098^{8} m/s. what is the frequency of the vibration?

7. Air passing through 20 equally spaced holes in a rotating disk produces a 2480-Hz tone. What is the angular speed of the disk?

8. Find the wavelength of a water wave of frequency 40 Hz traveling at 120 cm/s.

9. How long should an antenna be for a receiver with frequency of 30 MHz?

10. A 3 m string fixed at both ends vibrates in 4 distinct segments when driven by a 180 Hz source. What is the fundamental frequency of the string?

1. A police car is moving at 29 m/sand is behind a speeding car moving faster in the same direction as the police car. The radar gun in the police car emits an 8.0x10^{9} Hz electromagnetic wave. The frequency of the returning wave is measured to be 318Hz less than the emitted frequency. Find the speeder's speed.

2. The passenger in a car traveling 8 m/s judges the car's horn to have a frequency of 600Hz. If the sound of the horn gets reflected from a building ahead, (a) what is the frequency of the echo the passenger hears? (b) What beat frequency will the passenger hear?

3. A stationary observer hears a frequency of 560 Hz from an approaching car. After the car passes, the observed frequency is 460 Hz. What is the speed of the car? (speed of sound in air = 343 m/s)

4. A car approaching a stationary observer emits 450 Hz from its horn. If the observer detects a frequency of 470 Hz, how fast is the car moving?

Selected solutions are printed below.

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A6.

The total mechanical energy of the system is

E = K + U = (1/2)mω^{2}A^{2}

Since K = U

(1/2)mv^{2}+ (1/2)mv^{2} = (1/2)mω^{2}A^{2}

2v^{2} = ω^{2}A^{2}

v = ωA / √2

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B8.

The universal wave equation informs us

speed = frequency * wavelength

120 cm/s = (40 cycles/s)* wavelength

wavelength = 3.0 cm

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