A simple pendulum completes 10 oscillations in 25 seconds. Another simple pendulum completes 11 oscillations in same time. Compare the lengths of two simple pendulums.
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A simple pendulum makes 10 oscillations in 20 seconds. ... Time period=2 sec. Frequency =1/T=1/2=0.5 Hz ∴Time period and frequency of its oscillation are 2sec and 0.5 hertz. 4.4.
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Answered by
38
your answer is ---
°•° A simple pendulum completes 10 oscillation in 25 second
So, it's time period is T1 = 10/25 sec
•°•
T1 = 2π√L1/g
=> T1^2 = 4π^2 ×L1/g
=>( 10/25 )^3= 4π^2×L1/g
=> (100/625×g)/4π^2 = L1
°•° another pendulum complete 11 oscillation in 25 second
So, it's time period is T2 = 11/25 second
•°•
T2 = 2π√L2/g
=> (11/25)^2 = 4π^2×L2^2/g
=> (121/625×g)/4π^2 = L2
•°•
L1/L2 = 100/121
°•° A simple pendulum completes 10 oscillation in 25 second
So, it's time period is T1 = 10/25 sec
•°•
T1 = 2π√L1/g
=> T1^2 = 4π^2 ×L1/g
=>( 10/25 )^3= 4π^2×L1/g
=> (100/625×g)/4π^2 = L1
°•° another pendulum complete 11 oscillation in 25 second
So, it's time period is T2 = 11/25 second
•°•
T2 = 2π√L2/g
=> (11/25)^2 = 4π^2×L2^2/g
=> (121/625×g)/4π^2 = L2
•°•
L1/L2 = 100/121
Answered by
4
Answer
your answer is ---
°•° A simple pendulum completes 10 oscillation in 25 second
So, it's time period is T1 = 10/25 sec
•°•
T1 = 2π√L1/g
=> T1^2 = 4π^2 ×L1/g
=>( 10/25 )^3= 4π^2×L1/g
=> (100/625×g)/4π^2 = L1
°•° another pendulum complete 11 oscillation in 25 second
So, it's time period is T2 = 11/25 second
•°•
T2 = 2π√L2/g
=> (11/25)^2 = 4π^2×L2^2/g
=> (121/625×g)/4π^2 = L2
•°•
L1/L2 = 100/121
your answer is ---
°•° A simple pendulum completes 10 oscillation in 25 second
So, it's time period is T1 = 10/25 sec
•°•
T1 = 2π√L1/g
=> T1^2 = 4π^2 ×L1/g
=>( 10/25 )^3= 4π^2×L1/g
=> (100/625×g)/4π^2 = L1
°•° another pendulum complete 11 oscillation in 25 second
So, it's time period is T2 = 11/25 second
•°•
T2 = 2π√L2/g
=> (11/25)^2 = 4π^2×L2^2/g
=> (121/625×g)/4π^2 = L2
•°•
L1/L2 = 100/121
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