A simple harmonic oscillator consists of a block of mass 1.92 kg attached to a s

Question

A simple harmonic oscillator consists of a block of mass 1.92 kg attached to a spring of spring constant 100 N/m. When t = 1.00 s, the position and velocity of the block are x = 0.123 m and v = 3.417 m/s. What is the amplitude of the oscillations?

4.89×10-1 m You are correct. Your receipt no. is 162-6480 Help: Receipt Previous Tries

What was the position of the block at t = 0 s? Incorrect. Tries 10/13 Previous Tries

What was the velocity of the block at t = 0 s? Incorrect. Tries 2/13 Previous Tries

NOTE: The issue with this problem is that the answer for the final two questions needs to be reduced by 2*pi to obtain the phi value under one revolution

Explanation / Answer

Calculated by you that

Amplitude = 0.489 m

Now equation of wave will be

x = A*cos (wt + phi)

V = -w*A*sin (wt + phi)

w = sqrt (k/m) = sqrt (100/1.92)

w = 7.22 rad/sec

when t = 1.00 sec, then x = 0.123 m and V = 3.417 m/sec

dividing both bolded equation

V/x = -w*tan (wt + phi)

tan (wt + phi) = -V/(x*w)

wt + phi = arctan (-V/x*w)

phi = -w*t - arctan (V/x*w)

phi = -7.22*1.00 - arctan (3.417/(7.22*0.123))

phi = -7.22 - 1.3165

phi = -8.5365 rad

Now wave equation will be

x = 0.489*cos (7.22*t - 8.5365)

B.

at t = 0

x = 0.489*cos (-8.5365 rad) = -0.3084 rad

C.

V = -7.22*0.489*sin (7.22*t - 8.5365)

at t = 0

V = -7.22*0.489*sin (-8.5365 rad)

V = 2.739 m/sec

Comment below if you have any doubt.

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