8. Write the equation for simple harmonic motion given that the amplitude is 5 centimeters and the frequency is 2/ ð cycles per second. Assume that the maximum displacement occurs when t = 0.
a. y = 5sin4t
b. y = 5cos2t
c. y = 5cos4t
d. y = 5sin2t
9. Write the equation for simple harmonic motion given that the amplitude is 4 centimeters and the period is ð seconds. Assume zero displacement at t = 0.
a. y = 4cos2ðt
b. y = 4sin2ðt
c. y = 4cos2t
d. y = 4sin2t
*How would go about solving these? Need Help.
amplitude is pretty simple, that is the coefficent in front of the sin/cos term.
If it starts at zero amlitude at zero time, then it has to be sine
If it is maximum amplitude at zero time, it has to be cosine. Think out that.
Now the period. the standard format of the trig arguement is (2PI*frequency*time) I don't see PI in any of the answers, so I really don't understand.
on 8, with what I have given you, it has to be cosine, and the argument in the cosine term is 2PI*frequency*t
in 9, it has to be sine, with the argument 2PI*frequency*time.
To solve these questions, we need to understand the equation for simple harmonic motion:
y = A sin(ωt) or y = A cos(ωt)
where:
- y is the displacement from the equilibrium position,
- A is the amplitude (the maximum displacement from the equilibrium position),
- ω is the angular frequency (equal to 2πf), and
- t is the time.
Let's solve each question step-by-step:
8. Given the amplitude is 5 cm and the frequency is 2/π cycles per second, we can find the angular frequency as follows:
ω = 2πf = 2π(2/π) = 4 rad/s
Now, substitute the values into the equation:
y = 5 sin(4t)
Therefore, the answer is (a) y = 5 sin(4t).
9. Given the amplitude is 4 cm and the period is π seconds, we can find the angular frequency as follows:
ω = 2π/T = 2π/π = 2 rad/s
Now, substitute the values into the equation:
y = 4 sin(2πt)
Therefore, the answer is (b) y = 4 sin(2πt).
These are the steps to solve the given equations for simple harmonic motion.
To solve these questions, we need to understand the equation for simple harmonic motion (SHM). The general equation for SHM is given by:
y = A sin(ωt + φ)
Where:
- y is the displacement from the equilibrium position,
- A is the amplitude (maximum displacement),
- ω is the angular frequency (2πf), where f is the frequency in cycles per second,
- t is the time, and
- φ is the phase constant.
Now let's solve each question:
8. The amplitude is given as 5 centimeters, and the frequency is given as 2/π cycles per second. To find the equation, we need to find ω first. ω = 2πf = 2π(2/π) = 4π. Thus, the equation becomes:
y = 5 sin(4πt + φ)
Since it is mentioned that the maximum displacement occurs at t = 0, we can assume φ = 0. Thus, the correct equation for simple harmonic motion in this case is:
y = 5 sin(4πt)
So, the answer is option (a): y = 5 sin(4t).
9. The amplitude is given as 4 centimeters, and the period is given as π seconds. Since the period is the reciprocal of the frequency (T = 1/f), the frequency for this case is 1/π. The angular frequency, ω, is 2πf = 2π(1/π) = 2π. Thus, the equation becomes:
y = 4 sin(2πt + φ)
Since it is mentioned that there is zero displacement at t = 0, we can assume φ = π/2. Thus, the correct equation for simple harmonic motion in this case is:
y = 4 sin(2πt + π/2)
So, the answer is option (b): y = 4 sin(2πt).
To solve these types of questions, always remember the general equation for simple harmonic motion and match the given information with the appropriate variables in the equation.