Question

A harmonic motion has an amplitude of 0.2 cm and a time period of 0.15 s .Find the maximum velocity and acceleration. Ans. 8.38 cm/s, 350.9 cm/s2 Hint Since harmonic motion we have X = 0.2sin??t, ? can be found from time period Differentiate to find velocity and acceleration . A harmonic motion has a frequency of 10 Hz and its maximum velocity is 4.57 m/s .Determine its amplitude, its period , and its maximum acceleration . Ans. 7.27cm , 0.1 s ,and 278.1 m/ s2 An accelerometer is used to measure vibration of system it indicates that the system is vibrating at 82 c/s with maximum acceleration of 50 m/ s2 .Find the amplitude of vibration . Ans .0.019 cm

          A harmonic motion has an amplitude of 0.2 cm and a
time period of 0.15 s .Find the maximum velocity and
acceleration.
Ans. 8.38 cm/s, 350.9 cm/s2
Hint Since harmonic motion we have X = 0.2sin??t, ? can be found
from time period Differentiate to find velocity and acceleration .
A harmonic motion has a frequency of 10 Hz and its
maximum velocity is 4.57 m/s .Determine its amplitude,
its period , and its maximum acceleration .
Ans. 7.27cm , 0.1 s ,and 278.1 m/ s2
An accelerometer is used to measure vibration of system
it indicates that the system is vibrating at 82 c/s with
maximum acceleration of 50 m/ s2 .Find the amplitude
of vibration . Ans .0.019 cm

Show more…

A harmonic motion has an amplitude of 0.2 cm and a
time period of 0.15 s .Find the maximum velocity and
acceleration.
Ans. 8.38 cm/s, 350.9 cm/s2
Hint Since harmonic motion we have X = 0.2sin??t, ? can be found
from time period Differentiate to find velocity and acceleration .
A harmonic motion has a frequency of 10 Hz and its
maximum velocity is 4.57 m/s .Determine its amplitude,
its period , and its maximum acceleration .
Ans. 7.27cm , 0.1 s ,and 278.1 m/ s2
An accelerometer is used to measure vibration of system
it indicates that the system is vibrating at 82 c/s with
maximum acceleration of 50 m/ s2 .Find the amplitude
of vibration . Ans .0.019 cm

Added by Martin G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

07/07/2023

Step 1

- Amplitude \( A = 0.2 \) cm - Time period \( T = 0.15 \) s Show more…

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Transcript

00:01 Hello students, in this question given as amplitude of the harmonic oscillator is equal to 0 .2 cm and the time period is equal to 0 .15 seconds.

00:14 So, the angular velocity will be equal to 2 pi upon t.

00:19 Now, in this question we have to calculate the maximum velocity of the particle.

00:25 So, maximum velocity is equal to the formula is a into omega.

00:31 Now, we will substitute all the values 0 .2 cm into 2 into 3 .14 upon 0 .15.

00:40 So, after calculation we will get the maximum velocity of the particle is equal to the 8 .38 cm per second.

00:55 Next, we have to calculate the acceleration.

01:01 So, this is equal to minus a omega square.

01:05 The formula is this.

01:06 Now, we will substitute.

01:07 The negative sign shows that the particle is moving towards the center or the equilibrium position.

01:17 We can say that.

01:18 So, this is equal to minus 0 .2 into 2 into 3 .14 upon 0 .15 whole square.

01:30 So, after calculation we will get the acceleration.

01:32 After calculation, we will leave the negative sign only magnitude.

01:37 So, 350 .9 cm per second square.

01:48 In the next question given as frequency is equal to 10 hz...

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