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3 years ago

A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?

Physics

2 answers:

rjkz [21]3 years ago

8 0

Answer:

Maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec$

Maximum acceleration is given by $a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2$

So maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

Mnenie [13.5K]3 years ago

5 0

Answer:

1.11 m/s²

Explanation:

Amplitude, A = 0.43 m

Time period, T = 3.9 second

Let a be the maximum acceleration and ω be the angular frequency.

ω = 2π/T

ω = ( 2 x 31.4) / 3.9

ω = 1.61 rad/s

Maximum acceleration

a = ω²A

a = 1.61 x 1.61 x 0.43

a = 1.11 m/s²

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3 years ago

2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across

gulaghasi [49]

Answer:

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1 year ago

A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fir

bagirrra123 [75]

Answer:

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