An object vibrates in sipmple harmonic motion with time period 1.6 sec after passing the equilibrium position by 0.2 sec the speed of the object was 1.0m/s, what is the amplitude of the simple harmonic motion of the object? A)0.25m B)0.36m C)3.93m D)18.58

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dk_ch Share
Dec 8, 2017

Answer:

Option ( B)

Explanation:

drawn
Let the object #color(green)(P)# executes SHM on Y-axis maintaining its equilibrium position at origin O . Given that the time period of oscillation is #T=1.6#s. The angular velocity #omega# of the reference point #color(red) (R)# associated with SHM undergoing uniform circular motion will be given by #omega=(2pi)/T=(2pi)/1.6"rad/s"#.

Considering that time count is started when the object is at origin. So we can write the equation for displacement #x# in t sec as

#color(red)(y=asinomegat.....[1])# , where #a# is the amplitude of vibration

Here origin is also its equilibrium position and It returns repeatedly to this position after each complete oscillation for 1.6s.

So the velocity #v# of the object after #t # sec will be obtained by differentiating the above equation w r to t

#color(blue)(v=(dy)/(dt)=aomegacosomegat.......[2])#

Now it is given that the velocity of the object after passing the equilibrium position by #0.2#s is #1m"/s"#

So we can insert #t=0.2#s, #v=1m"/s"# in [2] to find out #a#

#v=aomegacosomegat#

#1=axx(2pi)/1.6xxcos((2pi)/1.6xx0.2)#

#=>1=axx(2pi)/1.6xxcos(pi/4)#

#=>1=axx(2pi)/1.6xx1/sqrt2#

#=>a=(0.8xxsqrt2)/pi~~0.36m#

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Dec 7, 2017

Answer:

(a) 0.25 m

Explanation:

Time period = 1.6 s
Angular frequency, #omega = (2π)/T = 3.927# rad s⁻¹

Since we know the object passes the equilibrium with a velocity of 1.0 m s⁻¹ that is the object's maximum velocity.

Use this equation to solve the problem:
#v_(max) = omega A#

#A = v_(max) / omega = 1.0/3.927 = 0.2546# m

So the answer is (a)

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