(a) The mechanical energy of the system is the sum of the potential energy and kinetic energy at any point in the oscillation. At maximum displacement from equilibrium (amplitude), all the energy is in the form of potential energy. At equilibrium position, all the energy is in the form of kinetic energy. Therefore, the mechanical energy of the system is equal to the potential energy at maximum displacement:
E = (1/2)kA^2
where A is the amplitude, k is the spring constant, and E is the mechanical energy.
Substituting given values, we get:
E = (1/2)(250 N/m)(0.035 m)^2 = 0.123 J
(b) The maximum speed of the block occurs at the equilibrium position, where all the energy is in the form of kinetic energy. Therefore, we can use the conservation of energy principle to find the maximum speed:
E = (1/2)mv^2
where m is the mass of the block and v is the maximum speed.
Substituting given values and solving for v, we get:
v = sqrt(2E/m) = sqrt(2(0.123 J)/(0.500 kg)) = 0.497 m/s
(c) The maximum acceleration of the block occurs at the extremes of the oscillation, where the net force on the block is maximum. The maximum net force is equal to the magnitude of the spring force at maximum displacement:
F = kA
where F is the net force, A is the amplitude, and k is the spring constant.
The maximum acceleration is equal to the net force divided by the mass of the block:
a = F/m = (kA)/m
Substituting given values and solving for a, we get:
a = (250 N/m)(0.035 m)/(0.500 kg) = 17.5 m/s^2
A block-spring system oscillates with an amplitude of 3.50 cm. The spring constant
is 250 N/m and the mass of the block is 0.500 kg. Determine
(a) the mechanical energy of the system,
(b) the maximum speed of the block,
(c) and the maximum acceleration
5 answers
M
I'm sorry, I'm not sure what you are asking. Could you please provide more information or context for your question?
A wave is modeled by the wave function y(x, t) = (0.30 m)sin �
2π
4.50 m (x − 18.00 m
s
t)
�
.
What are the
(a) amplitude,
1
(b) wavelength,
(c) wave speed,
(d) period,
(e) initial phase shift and
(f) frequency of the wave?
2π
4.50 m (x − 18.00 m
s
t)
�
.
What are the
(a) amplitude,
1
(b) wavelength,
(c) wave speed,
(d) period,
(e) initial phase shift and
(f) frequency of the wave?
(a) The amplitude is given as 0.30 m.
(b) The wavelength is related to the wave number, which is given by `k = 2π/λ`. Thus, the wavelength is given by `λ = 2π/k = 2π/4.50 m = 1.39 m`.
(c) The wave speed is related to the wavelength and the frequency, which are given in parts (b) and (f), respectively. The wave speed `v` is given by `v = λf`, so substituting the values we get `v = (1.39 m)(4.50 Hz) = 6.26 m/s`.
(d) The period is the inverse of the frequency, which is given in part (f). Thus, `T = 1/f = 1/4.50 Hz = 0.222 s`.
(e) The initial phase shift is given by the argument of the sine function, which is `-2π(18.00 m/s)t`. Therefore, the initial phase shift is `-2π(18.00 m/s) = -113.10 degrees` or `-1.98 radians`.
(f) The frequency is related to the angular frequency, which is given by `ω = 2πf`. Thus, `f = ω/2π = (2π/1 s)(4.50 m^-1) = 4.50 Hz`.
(b) The wavelength is related to the wave number, which is given by `k = 2π/λ`. Thus, the wavelength is given by `λ = 2π/k = 2π/4.50 m = 1.39 m`.
(c) The wave speed is related to the wavelength and the frequency, which are given in parts (b) and (f), respectively. The wave speed `v` is given by `v = λf`, so substituting the values we get `v = (1.39 m)(4.50 Hz) = 6.26 m/s`.
(d) The period is the inverse of the frequency, which is given in part (f). Thus, `T = 1/f = 1/4.50 Hz = 0.222 s`.
(e) The initial phase shift is given by the argument of the sine function, which is `-2π(18.00 m/s)t`. Therefore, the initial phase shift is `-2π(18.00 m/s) = -113.10 degrees` or `-1.98 radians`.
(f) The frequency is related to the angular frequency, which is given by `ω = 2πf`. Thus, `f = ω/2π = (2π/1 s)(4.50 m^-1) = 4.50 Hz`.