A quartz crystal of thickness 0.001m is vibrating at resonance. Calculate the fundamental frequency. Density of quartz = 2.650×103 kg/m3 and Young’s modulus for quartz = 7.9×1010 N/m2. a) 2.7299×106MHz b) 26.50×103Hz c) 2.093×106Hz d) 2.7299×1010MHz
Q: A 1.59-m-long rope is stretched between two supports with a tension that makes the transverse waves…
A: Given, Length of rope = 1.59m Speed of transverse wave = 64m/s We need to find, Wavelength and…
Q: A 2.30-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is…
A: Length (L) = 2.30 m Mass (m) = 0.100 kg Tension (T) = 18 N
Q: A taut string for which 0.054 kg/m is under a tension of 80 N. We want to generate sinusoidal waves…
A: The wave speed in string is calculated by using the formula v=Tμ , where T=80 N is the tension force…
Q: 2. A 3.00-m-long wire having a mass of 0.200 kg is fixed at both ends. The tension in the wire…
A:
Q: A piano wire with a length of .7m has a mass of 4.3×10^-3kg. What tension is needed for the wire to…
A:
Q: 1. Find the rms amplitude of the periodic signal with period T=2s shown in the following Figure. The…
A: Given : A signal with profile v= 2 e-2tOur task is to calculate the value of the rms value of the…
Q: Consider the experimental setup shown below. The length of the string between the string vibrator…
A: Step 1:Determine the given variables:The hanging mass is m=2 kg.The length of the string is L=3.0…
Q: 1.8 m string with a linear density of 18.5 x 10-3 Kg/m is stretched by a 225 N force. The string is…
A: Given: The length of the string is 1.8 m. The mass per unit length is 18.5x10-3 kg/m. The tension in…
Q: A lead cable of radius 1.6 cm is fixed at both ends and has a tension of 45,800 N. For this cable,…
A: Radius of cable R= 1.6cm Tension in cable T=45800N 10th harmonic=910Hz Density of cable ρ=11.3g/cm3…
Q: 7. A violin wire has a mass of 2.02 g and length 72.5 cm. It is stretched with a tension of 23.1 N.…
A: We have given Mass m = 2.02g = 2.02 × 10-3 kg Length L= 72.5 cm = 0.725 m Tension T = 23.1 N…
Q: A 5.50 g mass is suspended from a cylindrical sample of collagen 3.50 cm long and 2.00 mm in…
A: The expression for the Young’s modulus is,
Q: A 75.0 cm wire of mass 9.40 g is tied at both ends and adjusted to a tension of 39.0 N. You may want…
A: Given Tension, T = 39.0 N Length of wire, L = 75.0 cm = 0.75 m Mass ,M = 9.40 g
Q: A 2.35-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is…
A: (b) Given: The distance of the node from one end is 0.470 m. Introduction: A node is a point along a…
Q: A vibrating rope at a fixed tip has a frequency of (5 Hz), amplitude (12 cm) and a transverse wave…
A:
Q: 10. Review. The elastic limit of a steel wire is 2.70 X 108 Pa. What is the maximum speed at which…
A: Given: elastic limit of steel = pressure, P = 2.70*108 Pa density of steel, ρ = 7.86*103 kg/m3 To…
Q: A piano wire with a length of 0.7m has a mass of 4.3×10^-3kg. What tension is needed for the wire to…
A:
Q: A string along which waves can travel is 2.63 m long and has a mass of 103 g. The tension in the…
A:
Q: Your experiments on a particular insulator indicate that at 20°C, the average speed of sound in the…
A: Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts…
Q: A particular piano wire, of length .42 m, is stretched with a tension 900 N. If the mass of the wire…
A: Given data The length wire is given as L = 0.42 m. The tension in the string is T = 900 N. The mass…
Q: 1. Consider the experimental setup shown below. The length of the string between the string vibrator…
A: L=1m μ=0.0065 kg/mm=6.00 kgn=6vs=343 m/s
Q: What is a vibrational frequency of the n=6 harmonic of a string of length 0.71 m strung at the…
A:
A quartz crystal of thickness 0.001m is vibrating at resonance. Calculate the fundamental frequency. Density of quartz = 2.650×103 kg/m3 and Young’s modulus for quartz = 7.9×1010 N/m2.
a) 2.7299×106MHz
b) 26.50×103Hz
c) 2.093×106Hz
d) 2.7299×1010MHz
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
- Earthquakes at fault lines in Earth's crust create seismic waves, which are longitudinal (P-waves) or transverse (S-waves). The P-waves have a speed of about 9 km/s. Estimate the average bulk modulus of Earth's crust given that the density of rock is about 2100 kg/m3. PaWhat are the parts of this traveling wave? y(x,t) = (9.00 m) sin( (6л m¯¹)x + (4л rad/s)t - n/8) 3. Amplitude 4. Angular wave number13 = 342 237. € X Your response differs from the correct answer by more than 10%. Double check your calculations. m (c) Suppose the wire is carbon steel with a density of 7.89 x 10³ kg/m³, a cross-sectional area A = 2.68 x 10-7 m², and an elastic limit of 2.50 x 108 Pa. Find the fundamental frequency if the wire is tightened to the elastic limit. Neglect any stretching of the wire (which would slightly reduce the mass per unit length). Hz EXERCISE HINTS: GETTING STARTED I I'M STUCK! (a) Find the fundamental frequency and second harmonic if the tension in the wire is increased to 118 N. (Assume the wire doesn't stretch or break.) ffundamental = Hz f2nd harmonic= Hz (b) Using a sound speed of 342 m/s, find the wavelengths of the sound waves produced. λ (larger) = 2.00 X Follow the example. Once you have computed the correct fundamental frequency, you should be able to determine the wavelengths of the sound waves. m > (smaller) = 1.00 X Follow the example. Once you have computed the…
- Earthquakes at fault lines in Earth's crust create seismic waves, which are longitudinal (P-waves) or transverse (S-waves). The P-waves have a speed of about 9 km/s. Estimate the average bulk modulus of Earth's crust given that the density of rock is about 2800 kg/m3. PaA steel wire in a piano has a length of 0.420 m and a mass of 5.300 ✕ 10−3 kg. To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C (fC = 261.6 Hz on the chromatic musical scale)? NThe air pressure variations in a sound wave cause the eardrum to vibrate. A) Find the maximum velocity of the eardrum for vibrations of amplitude 5.15 × 10−8 m at a frequency of 20.0 Hz. m/s B) Find the maximum acceleration of the eardrum for vibrations of amplitude 5.15 × 10−8 m at a frequency of 20.0 Hz.in m/s^2 C) Find the maximum velocity of the eardrum for vibrations of amplitude 5.15 × 10−8 m at a frequency of 20.0 kHz. in m/s D) Find the maximum acceleration of the eardrum for vibrations of amplitude 5.15 × 10−8 m at a frequency of 20.0 kHz. m/s^2
- A travelling sinusoidal wave has this equation: y(x,t)=0.0450m sin (25.12m^(-1)x. –37.88^(-1)t–0.523) Find the following: d. Angular frequency e. Frequency f. Phase AngleA string of mass 16.6 grams and length 0.74 meters and fixed at the ends is under tension 243 N. What is the vibrational frequency of n= 4 harmonic?There is a long steel wire of length 2.7 m where its upper end is attached to the ceiling. At the other end, a 56.2 kg object is suspended. It is observed that it takes a transverse pulse 0.0345 s to travel from the bottom to the top of the wire. What is the mass of the wire? NOTE: Final answer in THREE decimal places. Include the unit. Round your answer to 3 decimal places.
- Chapter 35, Problem 029 Two waves of the same frequency have amplitudes 0.843 and 2.41. They interfere at a point where their phase difference is 79.0°. What is the resultant amplitude? Number Units Use correct number of significant digits; the tolerance is +/-2%4. (a) Sketch the pattern of vibration of a wire when it vibrates at twice its fundamental frequency. Label a node and an antinode on your diagram. (b) A horizontal string, rigidly held at its ends, is stretched between two points 0.80 m apart. The tension in the string is 130 N and its mass is 2.0 g. Calculate the mass per unit length of the string, the frequency of the fundamental mode of vibration of the string, and the speed at which a transverse wave travels along this string.