Any body moving with simple harmonic motion is being acted on by a force that is

Question

Any body moving with simple harmonic motion is being acted on by a force that is A string fixed at both ends is vibrating in a standing wave. There are three nodes between the ends of the string (not including the ends). The string is vibrating at a frequency that is its. An object (red arrow) is placed in front of a converting lens, as shown in the figure. In the figure, F marks the focal points, and 2F marks a distance twice the focal length. The image formed is In a single slit experiment if the width of the slit is increased, the width of the central maximum with If a negative charge were placed at the origin (the crossing point on the vertical and horizontal lines) of the figure, into which quadrant would it feel a net force?

Explanation / Answer

1)for simple harmonic motion, restoring force is proportional to the displacement and acts in opposite direction

for a mass connected to spring, force F=-k*x

where k=spring constant

x=displacement

hence option d is correct.

2)in total there are 5 nodes (including 2 nodes at the two ends)

hence total 4 segments.

hence its fourth harmonic.

(for fundamental, one segment between two end nodes exist, for n th harmonic, n segments exist)

Q3.

cartesian sign convention is used.

let object distance be u=-x*F.

here 1<x<2 as object lies within F and 2F.

focal length for converging lens is positive.

if image distance is v,

then (1/v)-(1/u)=1/F

==>(1/v)+(1/xF)=1/F

==>1/v=(x-1)/(x*F)

==>v=x*F/(x-1)

for x<2, x/(x-1)>2

hence v>2*F

hence the image is placed beyond 2F in right to the lens

hence it is a real image.

magnification=v/u=-1/(x-1)

as 1<x<2, 0<x<1

hence magnitude of v/u is greater than 1.

hence image is inverted and larger as compared to object.

option d is correct.

Q4.
width of central maxima is inversely proptional to slit width

hence if width is inccreased, maxima width will reduce

hence option b is correct

Q5.

force between two charged particles q1 and q2 ,separated by a distance is given by k*q1*q2/d^2

where k=9*10^9

if charges are of same sign, the force is repulsive in nature and if charges are of opposite sign, the force is attractive in nature

as all the four charges are positive and a negative charge is placed at origin, force is attractive in nature i.e. force due to each charge will be directed towards that charge.

as distance is constant, force magnitude will depend upon the magnitude of charge.

so charge having higher magnitude will have higher force

hence along y axis, net force will be directed along +ve y axis as 4q>2q

along x axis , net force is along +ve x axis as 3q>q

hence the resultant force will be in B quadrant

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