The force applied on the spring reduces hence the tension. The spring therefore accelerate mass faster, therefore the period would be shorter. b. Taking it to the moon where gravity is weaker. (U) The gravity change would have no effect on the period taken since the mass and spring are still the same, therefore no change is expected.
c. Weakening the spring (reducing the spring constant). (L) For a weakened spring, the force the spring exerts is decreased. The oscillations period would be therefore lengthened, would be longer d. Making the amplitude of the oscillation larger.(U) The amplitude does not affect frequency since the distance from relaxation position would increase the restoring force. Therefore, the frequency remains unchanged.
2. For a pendulum as in Fig 9.1.2: label how the following changes would affect the oscillation period. Label each as making the period shorter (S), longer (L), or unchanged (U). Explain your response. a. Taking it to a planet where gravity is larger.(S) Gravity affects the oscillation period from the formula of finding period using length and gravity. Therefore, as the gravity increases, the period decreases as they are inversely proportional b. Increase the mass hanging on the pendulum.(U) Period is mass independent. Therefore, at gravity all masses accelerate equally, hence the period is unchanged. c. Making the pendulum shorter. (S) The length is directly proportional to the period. Therefore a decrease in the pendulum length decreases the period d. Reducing the amplitude of the oscillation (assuming that it was not very big to start with). (U) The oscillations period remains constant due to the lack of relation to the amplitude. 3. The frequency of the tone produced by a violin string is higher (H), lower (L) or unchanged (U) if we make the following changes (note that here we are asking about the frequency, whereas on the earlier problems we were asking about the period of the oscillation, which is just the inverse of the frequency) : a. Making the string shorter. (H) The frequency of the tone is high. The shorter the string the higher the pitch, therefore the high frequency experienced. b. Making the string thicker. (L) The increased thickness increases the mass per unit length. Therefore the string moves slower which decreases the pitch, hence the frequency. c. Pressing the string down on the fingerboard.(H) The vibration reduces when the spring is pressed to the fingerboard; the active part is shortened. Therefore the pitch and frequency rose. d. Reducing the tension of the string. (L) The reduced tension of string causes slow movement of the string therefore the pitch and frequency reduced Explain your response. 4. I take a violin and make an exact copy of it, except that it is bigger. The strings are identical except for the length; they have the same material and the same tension. If the new violin is 2.30 times the size of the original, at what frequency would the string that was previously the A4 string (that is 440 Hz on a regular violin) oscillate? Use units of "Hz." Explain your response. When the size increases the pitch decreases, therefore 440Hz divided by 2.3 440Hz / 2.3= 191.30Hz 5. If your hearing cuts off at 17440 Hz, what is the highest harmonic of E5 string you can hear? The answer is an integer without units. Hint: The E5 string vibrates at 660 Hz. Explain your response. The highest harmonic is 17440Hz divided by 660Hz 17440/660=26.42 Rounding off, the highest harmonic to be heard is the 26th Harmonic 6. The frequency of the sound coming from the organ pipe is higher (H), lower (L) or the same (S) if we make the following changes to the organ: a. Moving the organ to a higher elevation. (H) The air is less dense at higher elevation, therefore the molecules move more