A 0.45 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by
x = (12 cm)cos[(17 rad/s)t + p/2 rad]
(a) What is the oscillation frequency (in Hz)? (b) What is the maximum speed acquired by the block? (c) At what value of x does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e) At what positive value of x does this occur? (f) What force, applied to the block by the spring, results in the given oscillation?
Give your answers in centimeter-based units, where applicable.

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kjuli1766 kjuli1766 · Answer 1 · 2022-12-19T12:31:20+01:00

(a) The angular frequency is 2.7 Hz

(b) The maximum speed is 2.04 m/s

(c) The maximum value is occur at mean position

so the dispalcement is zero

(d) The maximum acceleration is 34.68 m/s^2.

(e) The acceleartion should be maximum at extrem positions

so the position x = A = 12 cm

( f ) The force on the block 15.6 N

Calculation :

The mass of the block m = 0.45 kg

The dispalcement of the block

x = 12cm cos [(17 rad/s) t + (π/2 rad)]

(a) The angular frequency

ω = 17 rad/s

2πf = 17 rad/s

then the frequency

f = 17 / 2π = 2.7 Hz

(b) The maximum speed

v max = Aω = (12 cm) (17 rad/s)

= 2.04 m/s

(c) The maximum value is occur at mean position

so the dispalcement is zero

(d) The maximum acceleration

a max = Aω^2

= (0.12 m) (17)^2 = 34.68 m/s^2

(e) The acceleartion should be maximum at extrem positions

so the position x = A = 12 cm

(f) The force on the block

F = ma = (0.45 kg) (34.68 )

= 15.6 N

Force or Energy Exercised or Applied: Cause of Movement or Change: Active Force. force of nature

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