Q1 - Two identical blocks are attached to identical springs. The oscillation amp

Question

Q1 - Two identical blocks are attached to identical springs. The oscillation amplitude of Block A is 10cm; and the amplitude of Block B is 5 cm. Which block has a greater oscillation period?

Block A

Block B

The periods are the same for both Blocks.

Q2 - For an object in oscillation, which of the following is defined to be the maximum displacement from the equilibrium position.

Period

Frequency

Angular frequency

Amplitude

Q3 - A simple pendulum's period is 6.51 seconds . What is the angular frequency in rad/s?

Q5 - Which of the sensors must be zeroed before making measurement in this experiment?

Just motion sensor.

Just force sensor

Both motion and force sensor

Motion sensor, force sensor and meter stick

Block A

Block B

The periods are the same for both Blocks.

Explanation / Answer

Q1. C

given two identical blocks are attached to identical springs

and amplitude of Block A is 10cm; and the amplitude of Block B is 5 cm

we know that the time period is T = 2pi(sqrt(m/k))

so in the expression of Time period there is no amplitude term so the time period does not depend on amplitude.

answer is C The periods are the same for both Blocks.

Q2 D

For an object in oscillation,the maximum displacement from the equilibrium position defined to be the Amplitude

answer is option D

Q3

T = 6.51 s

the angular frequency is w = 2pi/T = 2pi/6.51 raad/s = 0.965159 rad/s

Q4

angular frequency W = 2.02 /s , amplitude A = 0.13 m

the magnitude of its maximun acceleration is a = ?

we know that F = -k*x

for maximum amplitude F = -kA = m*a

= - m*w^2*A = ma ==> a = W^2*A

the maximum acceleration is a = 2.02^2*0.13 m/s2 = 0.530452 m/s2

Q5

force sensor should be zeroed

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