For a simple pendulum to have a 1-second tick, what should be its length?

For a simple pendulum to have a 1-second tick, what should be its length?

• A. 0.5 m
• B. 1 m
• C. 1.5 m
• D. 2 m

Answer: (B)

For a simple pendulum to have a 1-second tick, its length can be determined using the formula for the period of a simple pendulum, which is given by: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on the surface of the Earth). Setting \( T \) to 1 second for a 1-second tick allows us to rearrange the formula to solve for \( L \): \[ 1 = 2\pi \sqrt{\frac{L}{9.81}} \] Squaring both sides gives: \[ 1 = 4\pi^2 \frac{L}{9.81} \] From this, we can isolate \( L \): \[ L = \frac{9.81}{4\pi^2} \] Calculating this yields approximately \( 0.248 \, \text{m} \) or \( 0.25 \, \text{m} \

For a simple pendulum to have a 1-second tick, its length can be determined using the formula for the period of a simple pendulum, which is given by:

[ T = 2\pi \sqrt{\frac{L}{g}} ]

where ( T ) is the period, ( L ) is the length of the pendulum, and ( g ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 ) on the surface of the Earth).

Setting ( T ) to 1 second for a 1-second tick allows us to rearrange the formula to solve for ( L ):

[ 1 = 2\pi \sqrt{\frac{L}{9.81}} ]

Squaring both sides gives:

[ 1 = 4\pi^2 \frac{L}{9.81} ]

From this, we can isolate ( L ):

[ L = \frac{9.81}{4\pi^2} ]

Calculating this yields approximately ( 0.248 , \text{m} ) or ( 0.25 , \text{m} \