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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Simple Harmonic Motion
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rhd92781 (686)

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1) in fig, the block of mass m, attached to the spring of stiffness k is in contact with the completely elastic wall, and the compression in the spring is 'e'. The spring is compressed further by 'e' by displacing the block towards left and is then released. If the collision b/w the block and the wall is completely elastic then the time period of oscillations of the block will be:
(a) 2(pi)/3 root(m/k)
(b) 2(pi) root(m/k)
(c) pi/3 root(m/k)
(d) pi/6 root(m/k)
ans: a
(fg attchd)
 
 
2) One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross-section and the Young's modulus of the wire are A and Y respectively. if the mass is slightly pulled down and released, it will oscillate with a time period T equal to:
ans: 2(pi) root(m(YA + KL)/YAK)


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rahulknk (0)

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my ans is coming b
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priyata (5)

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W(angular valocity)
w=root(k/m)
gen. eqn of motion:
x=2ecoswt;
e=2ecoswt
wt=pi/3
t*root(k/m)=pi/3
t=pi/3*root(m/k)
T=2t=2pi/3root(m/k)
option a

EVERYONE MAY HAVE THE X-FACTOR.
BUT....................
I HAVE THE X+dx ONE.
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rhd92781 (686)

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how did u get the equation
e=2ecos(wt)
because when t=0, x=-2e

<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>

I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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