Answer :
The mass of the pendulum does not affect the period and frequency
The period can be determined using the formula:
T = 2π √(L/g)
where L is the length of the rope
and g is the acceleration due to gravity = 9.8 m/s2
Plugging in the values:
T = 2π √(5/9.8)
T = 10π/7 = 4.49 s
And for the frequency:
f = 1/T
f = 1/4.49
f = 0.22 Hz
The period can be determined using the formula:
T = 2π √(L/g)
where L is the length of the rope
and g is the acceleration due to gravity = 9.8 m/s2
Plugging in the values:
T = 2π √(5/9.8)
T = 10π/7 = 4.49 s
And for the frequency:
f = 1/T
f = 1/4.49
f = 0.22 Hz
Answer:
Period = 4.5 seconds
Frequency = 0.22 hertz.
Step-by-step explanation:
Given : 10-pound weight oscillating (as a pendulum) on a 5-meter-long rope.
To find : What is the period and frequency.
Solution :
The formula to find time when length is given,
[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]
where,T is the time period , L is the length and g is the acceleration due to gravity
L= 5 m and g=9.8 m/s
Substitute in the formula,
[tex]T=2\pi\sqrt{\frac{5}{9.8}}[/tex]
[tex]T=2\pi\sqrt{0.510}[/tex]
[tex]T=2\times 3.14\times 0.714[/tex]
[tex]T=4.483[/tex]
The time period is approx 4.5 seconds.
[tex]\text{Frequency}=\frac{1}{\text{Period}}[/tex]
[tex]\text{Frequency}=\frac{1}{4.5}[/tex]
[tex]\text{Frequency}=0.22\text{hertz}[/tex]
Therefore, Period = 4.5 seconds
Frequency is 0.22 hertz.