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Sample Questions
Expert help when you need it- Q1:A 3.5 kg object subject to a restoring force F is undergoing simple harmonic motion with a small amplitude. The potential energy U of the object as a function of distance x from its equilibrium position is shown above. This particular object has a total energy E: of 0.58 J. 1. What is the farthest the object moves along the x-axis in the positive direction? Explain your reasoning. 2. Determine the objects kinetic energy when its displacement is +7 cm. 3. Determine the objects speed when its displacement is -4 cm. 4. What is the object's speed at x = 0?
Answeri) The farthest the object can move is the instant when kinetic energy becomes zero or potential energy U = 0.58 From graph, it is clear that potential energy will reach 0.58 J at 12 cm ii) At a displacement of 7 cm, from graph we have: U = 0.2 J But we have: E=U+K 0.58=0.2+K K =0.38J From this, we have: , K =-1 mv-2 0.38= 1 —*3.5*v" 2 v= 0.466— s iii) At a displacement of —4 cm, from graph we have: U = 0.05 J But we have: E=U+K 0.58=0.05+K K = 0.53J From this, we have: K =-1 Pnv-, 2 0.53 =-1 *3.5*v2 2 v=0.55'11 iv) Total Energy is given as E = 0.58 J At x = 0, from graph we see that potential energy U = 0 Therefore, total energy will be same as kinetic energy K =-1/2 mv^2 0.58 = 0.5*3.5* v^2 v= 0.5757 m/ s - Q2:Suppose you are in a car, and you start from rest on a curve with radius r = 120 m and accelerate tangentially at 1.0 m/s^2. We would like to know the position, speed, and acceleration of the car after 2.0 seconds.Answer1. Calculate the angular acceleration: Angular acceleration = Tangential acceleration / Radius Angular acceleration = 1.0 m/s^2 / 120m Angular acceleration = 0.0083 rad/s^2 2. Calculate the angular velocity after 2.0 seconds: Angular velocity = Initial angular velocity + (Angular acceleration * time) Angular velocity = 0 + (0.0083 rad/s^2 * 2s) Angular velocity = 0.0166 rad/s 3. Calculate the linear velocity using the formula: Linear velocity = Radius * Angular velocity Linear velocity = 120m * 0.0166 rad/s Linear velocity = 1.992 m/s 4. Calculate the linear acceleration using radial and tangential components: Radial acceleration = (Linear velocity)^2 / Radius Radial acceleration = (1.992 m/s)^2 / 120m Radial acceleration = 0.033 m/s^2 Total acceleration = √ (Tangential acceleration)^2 + (Radial acceleration)^2 Total acceleration = √ (1.0 m/s^2)^2 + (0.033 m/s^2)^2 Total acceleration = √1.001 m/s^2 Total acceleration ≈ 1.00 m/s^2 Therefore, after 2.0 seconds, the car's position, speed, and acceleration on the curve with radius r = 120 m and tangential acceleration of 1.0 m/s^2 would be: - Position: Determined by the arc length traveled - Speed: 1.992 m/s - Acceleration: Approximately 1.00 m/s^2
- Q3:Complete the table below for a waves on a particular string of length L. All waves travel at the speed v, based on the tension and density of the string.
AnswerCertainly! To calculate the wavelength (λ) and frequency (f) for each mode in terms of the string length ( L ) and the speed of wave propagation ( v ), you can use the following formulas:
For the ( n^{th} ) mode:
Wavelength (λ): λn =2L/n
Frequency (f): fn =n⋅v/2L
Here’s how you can express the wavelength and frequency for each mode:
Mode 1:
Wavelength: λ=2L/1=2L
Frequency: f=1⋅v/2L=v/2L
Mode 2:
Wavelength: λ=2L/2=L
Frequency: f=2⋅v/2L=Lv
Mode 3:
Wavelength: λ=2L/3
Frequency: f=3⋅v/2L
Mode 4:
Wavelength: λ=2L/4=L/2
Frequency: f=4⋅v/2L=2v/L
These expressions give you the wavelength and frequency for each mode as a function of the string length ( L ) and the speed of wave propagation ( v )
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