A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.70 Hz. At $t = 0$ s, the mass is at $x = 4.40$ cm and has $v_x = -26.0$ cm/s. Part H Determine the position at $t = 4.80$ s. Express your answer with the appropriate units. $x = -4.18$ cm
Added by Anna S.
Close
Step 1
70Hz. Given f = 3.70Hz, we can rearrange the formula to solve for ω: ω = 2π * f ω = 2π * 3.70Hz ω ≈ 23.25 rad/s Show more…
Show all steps
Your feedback will help us improve your experience
Sahil Kumar and 101 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Question: A mass is pulled 19.3 cm below its equilibrium point and released at t = 0 s. This causes it to undergo simple harmonic motion with an angular frequency 5.70 rad s!. Suppose the mass is now doubled and then is pulled down by the same amount and released at t = 0 s. Remember to use sqrt(x) to enter √̅x̅. Part 1) How does the new period relate to the old period? Tₙₑဠ = [ ] Tₒₗₑ Part 2) Answer parts 2 and 3 with decimals correct to 3 significant figures. Write an expression for the displacement from equilibrium with time (the doubled mass remains on the spring). Take displacements below the equilibrium position as negative. Note: Please enter a positive answer for the cos( ) function. Even though cos(-x) = cos(x) STACK marks the negative answer as incorrect. x(t) = [ ] cos( [ ] ) cm Part 3) Write an expression for the velocity of the mass as a function of time. v(t) = [ ] sin( [ ] ) cm/s
Sahil K.
An ideal horizontal spring-mass system oscillates with an amplitude of 0.300 m. The spring has a constant of 100 N/m and it's observed that the mass is moving at 1.80 m/s when it is 0.100 m from its equilibrium position. As the mass oscillates, its period is Answer seconds. (answer with 3 sig figs) An ideal horizontal spring-mass system oscillates with an amplitude of 0.300 m. The spring has a constant of 100 N/m and it's observed that the mass is moving at 1.80 m/s when it is 0.100 m from its equilibrium position. As the mass oscillates, its maximum speed is Answer m/s. (answer with 3 sig figs)
Supreeta N.
A 1.30 kg mass on a spring has displacement as a function of time given by the equation x(t) = (7.40 cm) cos[(4.16 rad/s) t - 2.42 rad]. You may want to review. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Angular frequency, frequency, and period in shm. Part A Find the time for one complete vibration. Part B Find the force constant of the spring. Part C Find the maximum speed of the mass. Part D Find the maximum magnitude of force on the mass.
Ivan K.
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD