question 1) A mass of 0.25 kg is attached to a spring and is set into vibration

Question

question 1) A mass of 0.25 kg is attached to a spring and
is set into vibration with a period of 0.18 s.
What is the spring constant of the spring?
Answer in units of N/m.

question 2) A 67.2 g mass is attached to a horizontal
spring with a spring constant of 11.2 N/m
and released from rest with an amplitude of
27.4 cm.
What is the velocity of the mass when it
is halfway to the equilibrium position if the
surface is frictionless?
Answer in units of m/s.

question 3) You need to know the height of a tower, but
darkness obscures the ceiling. You note that
a pendulum extending from the ceiling almost
touches the floor and that its period is 15 s.
The acceleration of gravity is 9.81 m/s2 .
How tall is the tower?
Answer in units of m.

Explanation / Answer

1) Given that mass m = 0.25 kg time period is T = 0.18 s The time period is            T = 2[m/k]            k = 4^2 (m/T^2)               = 4^2 ((0.25 kg)/(0.18 s)^2)               = 304.617 N/m 2) Given that mass m = 67.2 x 10^-3 kg spring constant k = 11.2 N/m Amplitude is A = 0.274 m The velocity of the mass when it is halfway to the equilibrium position if the
surface is frictionless is        v = [(k/m)(A^2 - x^2)] Here x = A/2      v= [(k/m)(A^2 - (A/2)^2)]        = [(k/m)(3A^2/4 )]        =A [(k/m)(3/4 )]        = ( 0.274 m)[(( 11.2 N/m)/(67.2 x 10^-3 kg))(3/4 )]        = 3.063 m/s Note: Please submit one question per post as per guidelines. Note: Please submit one question per post as per guidelines.

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