PHYSICS Class
II IIT-JEE Achiever 2016-17 Intensive Revision Program Solution to Worksheet – 14
Topics
Simple Harmonic Motion
Date:
20-12-2016
Subjective Questions 2
1. A particle moves under the force F = (x − 6x) N, where x is in metres. For small displacements from the origin what is the force constant in the simple harmonic motion approximation? Solution 2
F = x − 6x For small displacement from the origin, the force is 2
2
F′ = F(x + ∆x) − F(x) = (x + ∆x) − 6(x + ∆x) − x + 6x = 2x ∆x − 6∆x = − (6 − 2x) ∆x = − 6∆x For x = 0 = k ∆x ⇒ k = 6 N/m A 2. At x = , what fraction of the mechanical energy is potential energy? What fraction is kinetic? Assume 2 potential energy to be zero at mean position. Solution 2
2
2
v=ω A −x =ω A −
A2
3
Aω 2 1 1 3 3 1 3 K.E. = mv2 = .m. A2 ω2 = . mA2 ω2 = (translational energy) 2 2 4 4 2 4 4
=
P.E. = Translational energy − K.E. 3 1 = Translational energy − (T.E.) = × (Translational energy) 4 4 3. A body executing SHM covers a, b in successive seconds starting from one end. Find its amplitude. Solution Since the body starts from one end, x = a cos ωt Since t = 1 and displacement OP = A −a Hence, x = A −a = A cos ω
…(1)
Where A is amplitude of the SHM Similarly, A −(a + b) = A cos 2 ω
⇒ From equation (1), A(1 − cos ω) = a 1 − cos 2 ω 1 − cos ω
=
2(1 − cos 2 ω)
=
a +b
= 2(1 + cos ω) ∴ cos ω =
1 − cos ω a a a 2a 2 2a 2 ∴A = = = = b − a 2a − b + a 3a 1 − cos ω 3a − b 1− 2a 2 2a ∴ Amplitude of the SHM = 3a − b 2IIT1617PPWS14S
1
a+b 2a
−1 =
a + b − 2a 2a
=
b−a 2a
4. A particle executing SHM moves from one end to the other. Its distances from the mid-point of its path, 2π at successive seconds, are observed to be x 1, x2, x3. Prove that its time period is x + x3 cos −1 1 2x 2 Solution As x = a sin ωt
∴ x1 = a sin ω; x2 = a sin 2ω, x3 = a sin 3ω x + x3 a[sin ω + sin 3ω] 2 sin 2ω. cos ω ∴ 1 = = = cos ω 2x 2 2a sin 2ω 2 sin 2ω ∴ ω = cos−1
x1 + x 3
∴ T=
2x 2
2π
ω
= cos
2π −1 x1 + x 3 2x 2 2
5. A body of mass 200 g is in equilibrium at x = 0 under the influence of a force F = ( −100 x + 10x )N. (a) If the body is displaced a small distance from equilibrium, what is the period of its oscillations? (b) If the amplitude is 4.0 cm, by how much do we error in assuming that F = −kx at the end points of
the
motion. Solution (a) T = 2π
m k
0.2
= 2π
100
=
2π
0.2 = 0.28 s
10
2
16 4 ) 100 × (b) (Fact ua + 10 × = −4 + ual max . = − 100 1000 100 4
(Factual)max. = − 4
−4 + % Error =
16 1000 4
+4 × 100% =
4 10
% = 0.4% 2
6. A point moves along the x-axis according to the equation x = a sin ωt − period, velocity and projection vx as a function of x. Solution
π π π x = a sin ωt − = a sin ωt cos − cos ω t sin 4 4 4
2
2
1
= a (sin ω t − cos ωt ) 2 ,
since
2
⇒x=
a 2
sin
π 4
= cos
π 4
1
=
2
(1 − sin 2ωt )
…(1)
∴ amplitude =
Maximum displacement occurs when sin 2ωt = 0 Period is given by T =
2π 2ω
or T =
a 2
π ω
2x = (+ 2ω cos 2ω t ) = + aω cos 2ωt = aω 1 − sin 2ω t = + aω 1 − 1 − dt 2 a
dx
a
Because from (1), 1 −
2IIT1617PPWS14S
2
2x a
= sin 2ωt
2
2
π . Find the amplitude, 4
dx
∴ vx =
dt
= aω 2 −
2x 2x
xx = 2aω 1 − = 2ω (a − x ) x a a a a
7. A particle of mass m is located in a unidimensional potential field where the potential energy of the particle depends on the coordinates as U(x) = U 0(1 − cos ax), U0 are a constants. Find the period of small oscillations that the particle performs about the equilibrium position. Solution U = P.E. = U0(1 − cos ax)
Ths is zero if cos ax = 1
∴ ax = 0 or x = 0 This is the mean position of the body executing SHM. At extreme position, P.E. is maximum. It is 2U0. 2U0 = U0(1 − cos ax) = U 0 (1 − cos Aa) Because at the extreme position, the displacement x = amplitude, A
∴ 2 = 1 − cos aA Since A is small in SHM, cos aA = 1 −
(Aa)
2
2
A2a 2 α is small, cos α = 1 − because when α is ∴ 1 − 1 − 2 2 2
2 2
Or A a = 4
or Aa = 2
or
Maximum P.E. = 2U0 = Maximum K.E. = 2
∴ω =
u0a m
2
or ω = a
1 2
A=
mω2 A 2 =
1 2
=2
2 a
mω2
4 a2
µ0 2π 2π m = ∴ Period of small oscillations = T = ω a U0 m 2
2
2
8. The speed v of a particle moving along x-axis is given by, v = 8bx − x − 12b , where b is a constant. Find amplitude of oscillations. Solution Amplitude of oscillations is the maximum separation of the particle from mean position to the extreme position. Also the speed of the particle becomes zero at extreme position. Let x represents extreme positions, then 2
2
2
2
8bx − x − 12b = 0 or x − 8bx + 12b = 0
or (x − 6b) (x − 2b) = 0
∴ x = 2b and 6b It shows that particle moves along x-axis from x = 2b to 6b If A is the amplitude of oscillations, then 2A = 6b − 2b = 4b
∴ A = 2b 9. A small mass m is fastened to a vertical wire which is under tension T as shown in the figure. What will be the frequency of vibration of the mass m if it is displaced laterally a slight distance and then released?
2IIT1617PPWS14S
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Solution Suppose the mass m is at a distance x to the right of the equilibrium position.
If the tension is T in the wire, the mass is acted upon by the horizontal components of the tension towards the left. These components are T cos α, T cos β .
∴ Restoring force = − (T cos α + T cos β )
x
= −T
2 2 x +b
+
x x = −T + 2 2 2 2 x +c x x c 1+ 2 b 1 + b2 c 1 − 2 2 x2 2 1 − x 2 1 − x 2 1 + 2 c ≈ − Tx 2b + 2c + c c b x
1 − 2 2 x 1+ b2 = − Tx b x2 x2 1 1 = − Tx + , ne neglecting , 3 terms since x is small. 3 b c 2 b 2c 1 1 −T + x b c Acceleration =
m
Thus, acceleration is proportional to displacement. So, the body executes S.H.M.
∴ Period = 2π
∴ Frequency =
m 1 1 b + c T
1 2π
T b + c Hz m bc
10. A uniform board of length L and weight W is balanced on a fixed semicircular cylinder of radius r as shown in the figure. If the plank is tilted slightly from its equilibrium position, determine its period of oscillation. Solution When tilted through a small angle θ, the point of contact shifts from B to B ′ through a small distance rθ. So, the torque on the rod about the point B Mg(BB′) = Mg.r sinθ = Mgrθ (as θ is small) d 2 θ ML2 d 2 θ This must be equal to I 2 = 2 dt 12 dt
2IIT1617PPWS14S
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ML2 rod abou bout its its centre tre = M.I. of the rod 12 ∴
ML2 d 2 θ 12 dt
2
= Mgrθ 2
∴ T = 2π
L
12gr
= πL
∴
d 2θ dt
2
=
12grθ 2
L
θ
1 3gr
11. A uniform cylindrical pulley of mass M and radius R can freely rotate about the horizontal axis O. The free end of a thread, lightly wound on the pulley, carries a block of weight A. At a certain angle α, it counterbalances a point mass m, fixed at the rim of the pulley. Find the frequency of small oscillations of the arrangement. Solution Initially, the torque is zero because mg(PK) = A × R
⇒ mg R sin α = A.R
∴ A = mg sin α
Let A be pulled down so that m goes up a little and α becomes (α + dα) Increase in torque = mgR sin ( α + dα) −mgR sin α Restoring torque = − [mgR (sin α cos dα + cos α sin dα) −mgR sin α] = − [mgR cos αdα + mgR sin α−mgR sin α] Because when dα is very small sin d α = dα, cos dα = 1
∴ Restoring torque = −mgR cos α dα MR 2 AR 2 R 2 α′ 2 ⇒ Iα′ = + + mR α′ = 2 g 2
A
2A + 2m M + g
where α′ is the angular acceleration. R 2α′ ∴ − mgR cos αdα = {M + 2m(1 + sin α)} × angular acceleration 2
∴ Angular displacement is proportional to angular acceleration. It executes SHM
∴ Period = T = 2π − Or frequency is v =
displacement acceleration
= 2π
R {M {M + 2 m(1 + sin α )} 2
mg cos α
2mgcos α MR + 2mR (1 + sin α )
12. A uniform board is placed on two spinning wheels as shown in figure. The axes of the wheels are separated by a distance l = 20 cm. The coefficient of friction between the board and the wheels is K = 0.18. Explain that in this case the board performs harmonic oscillations. Find the period of these oscillations.
Solution Each roller exerts on the board forces of friction equal to f 1 = KN 1 and f 2 = KN2, where N1, N2 are the normals of the board on the rollers. These force f 1, f 2 are shown in the figure. Suppose the centre of gravity of the board is slightly displaced. Let it be x from the middle. 2IIT1617PPWS14S
5
Then, the forces f 1, f 2 will not be equal because N1 and N2 will be different. l + x N1 = W l
where W is the weight of board and the distance between the wheel axes is l. l−x K (l + x ) K(l − x ) N2 = W ∴ f1 = W and f 2 = W l
l
∴ Resultant force =
K l
l
W [l + x − l + x ] =
∴ Acceleration of rod =
2KWx
W g
2KW l
x
, since mass of rod =
l
∴ Acceleration =
W g
2Kgx l
So, acceleration is proportional to displacement. So, it executes SHM.
∴ Period T = 2π
l
2 Kg
= 2π
20 2 × 100 × 0.18 × 9.8
= 2π
1 2 × 49 × 0.18
= 1.5 sec
13. A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period. Solution Let any instant, the body is at angular position θ with respect to the vertical line drawn from the centre of the mirror. If φ is the angular displacement of the ball about its centre, r ∴θ = then (R − r )θ = rφ φ − R r
Restoring torque acting on the ball, τ = − (mg sin θ)r For small θ, sin θ≈θ
∴ τ = −mgθ × r
r2 r φ = r m g ( −φ ) R − r R − r
or τ = − mg
2
Now comparing above equation with standard equation of SHM, α = −ω φ, we get ω =
mg r 2
I R −r
r2 5 g ∴ω= = 7 R−r 7 2 R−r 5 mr mg
2
(I = (7 mr /5) is the M.I. of the rolling ball about point of contact) contact)
and T =
2π
ω
=2
7( R − r ) 5g
14. A thin rod of length L and mass M is free to rotate in a vertical plane about a horizontal axis passing through the point O where one end of the rod is pivoted as shown in the figure. The other end of the rod 2IIT1617PPWS14S
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is connected to a vertical massless spring of force constant k. The lower end of the vertical spring is rigidly fixed to the ground at G. When the rod is in equilibrium position it is parallel to the ground. When the rod is slightly rotated from its equilibrium position and released. (a) Find its time period T of small oscillations. (b) What will be the maximum linear speed of the displaced end of the rod if the amplitude of oscillations is taken θ0?
Solution (a) The rod is initially horizontal and in the equilibrium position. When the end P of the rod is slightly compressed to angular displacement say θ, the spring compresses to y(say). The restoring force F = −ky acts at this end P in vertically upward direction. Thus, torque acting on the rod about O is
where tan θ ≈ θ =
τ = F.L = −kyL
⇒ y = Lθ
2
∴τ = − kL θ
…(2) (where α is the angular acceleration)
(and momentum of inertia of the rod about its axis of rotation or
d θ dt
2
+
3k M
2
θ=0
or
L
…(1)
Now by definition, 2 2 d θ 1 2 d θ τ = Iα = 2 = ML . 2 dt 3 dt 2 1 d θ ∴ ML2 . 2 = − kL2θ 3 dt
2
y
dθ dt
2
+ ω2 θ = 0 where ω =
Period of angular oscillations of the rod T =
2π
ω
= 2π
3k
1 3
ML2 )
M M 3k
(b) Maximum angular speed of the rod is in its equilibrium position
ωmax = ωθ0 (similarly as vmax = ωA) Therefore, maximum linear speed is vmax = Lωmax = ωLθ0 = Lθ0
3k M
15. A uniform rod of mass m and length l is suspended through a light wire of length l and torsional constant k as shown in the figure. Find the time period if t he system makes
2IIT1617PPWS14S
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(a) small oscillations in the vertical plane about the suspension point and (b) angular oscillations in the horizontal plane about the centre of the rod Solution (a) The oscillations take place about the horizontal line through the point of suspension and perpendicular to the plane of the figure. The moment of inertia of the rod about this line is
13ml 2 + ml = 12 12
ml 2
2
I
The time period = 2π
= 2π
mgl
13ml
2
12mgl
= 2π
13l 12g
(b) The angular oscillations take place about the suspension wire. 2 ml The moment of inertia about this line is 12 The time period is 2π
I k
= 2π
ml 2 12 k
Multiple choice questions with one correct alternative
1. Two cars A and B depart simultaneously from the same position and in same direction on a straight road. −1
−2
A starts with initial velocity 2 m s and acceleration 2 m s . While B starts with initial velocity 2 m s
−1
−2
and acceleration 4 m s . The driver of car A hears a sound of frequency 352 Hz emitted by car B after 10 s after the start. Find the actual frequency of the sound as emitted by B (Take velocity of the −1
sound = 330 m s ) (A) 271 Hz
(B) 371 Hz
(C) 550 Hz
(D) 713 Hz
Ans (B) Ans (B) Let t be the time at which sound is produced by B, then distance travelled by B, 1 s B = 2t + × 4 × t 2 and speed of B, v B = 2 + 4t 2 When car A, receives sound, distance travelled by A, 1 s A = 2 × 10 + × 2 × 10 2 = 120 m 2 −1
and speed of A, v A = 2 + 2 × 10 = 22 m s . Time taken for the sound to reach the car is (10 − t)
∴ Distance travelled by the sound = v × (10 − t) = 330(10 − t) 2 and distance between the cars = s B − sA = (2t + 2t) − 120 2 2t + 2t − 120 = 3300 − 330t
... (1) ... (2)
2
2t + 332t − 3420 = 0 On solving, solving, we get t = 9.37 s
∴ vB = 2 + 4 × t = 41 m s−1
v + v B v v + A 330 + 41 = 352 = 371 Hz 330 + 22
∴ apparent frequency heard by the driver of of car A is f ′ = f
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2. In a sonometer wire, the tension is maintained by suspending a 50.7 kg mass from the free end of the 3
wire. The suspended mass has a volume of 0.0075 m . The fundamental frequency of the wire is 260 Hz. If the suspended mass is completely submerged in water, the fundamental frequency will become (A) 200 Hz
(B) 220 Hz
(C) 230 Hz
(D) 240 Hz
Ans (A) Ans (A) Given f = 260 Hz, when T = 507 N. When mass is submerged, under water Upthrust = FB = Vρg 3
FB = 0.0075 × 10 × 10 = 75 N.
∴ Tension in the string, T / = 507 − 75 = 432 N f =
1
T
2l
µ
and f / = (2) f / (1)
f
… (1)
1
T /
2l
µ T /
=
T
… (2)
=
432 507
=
144 169
=
12 13
12 12 ⇒ f / = f = 260 × = 240 Hz. 13 13 3. The vibrating portion of a wire which is stretched with a weight of 6.48 kg weighs 0.5 g. When sounding in fundamental note, it is found to give 20 beats in 5 seconds, with vibrating tuning fork of frequency 256. If the length of the wire is slightly decreased, the note emitted by it is observed to be in unison with that of the fork. The original length of the wire is (A) 1 m
(B) 0.5 m
(C) 1.5 m
(D) 2 m
Ans (B) Ans (B) When the length of the wire is l metre, the number of beats per second =
20 5
=4
If the length is decreased, the frequency increase and the beats disappear. The original frequency of the wire is less than 256 by 4. So the frequency of the wire of length l is f = 256 − 4 = 252 Hz T = 6.48 × 9.8 N m=
0.5 × 10 −3
kg m −1
1
∴ 252 =
1 2l
or (252) 2 =
6.48 × 9.8 × L 0.5 × 10 −3 1 4l 2
×
6.48 × 9.8 × L 0.5 × 10 −3
⇒ l = 0.5 m = 50 cm
4. The time period of a particle in SHM is 8 seconds. At t = 0, it is at the mean position. The ratio of the distances traveled by it in the first and second is 1 1 (A) (B) 2 2
(C)
Ans (D) Ans (D) y1 = a sin ωt 2IIT1617PPWS14S
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1 3
(D)
1 2 −1
nd
For 1 sec, y 1 = a sin ω. y1 sin ω 1
For 2 sec, y2 = a sin 2 ω 1 = = = 2π y 2 sin 2ω 2 cos ω 2cos T y1 1 1 1 = = = ⇒ y 2 = 2y1 2π y2 1 2 2cos 2 8 2
(
nd
Distance covered in 2 sec = y2 − y1 =
)
2 − 1 y1
∴ Ratio =
1 2 −1
5. A vertical U-tube of uniform cross-section contains water upto a height of 30 cm. If the water on one side is depressed and then released, its motion up and down the two sides of the tube is simple harmonic. The time period of this SHM is nearly (A) 4 s
(B) 3 s
(C) 2 s
(D) 1 s
Ans (D) Ans (D) Figure shows a U-tube of uniform cross-sectional area A. Let the liquid be depressed through a distance y in one limb. Then the water rises through the same distance y in another limb. The difference of levels between two limbs will be 2y. Excess of pressure on whole liquid liqui d = (difference in height) (density) ( density) (g) = 2y × 1 × g
3
( ∵ density of water = 1 g / cm )
Force on the liquid = Pressure × area = 2 y g A Due to this force the liquid accelerates. The mass of of whole liquid in U tube = Volume × density
Acceleration, a = a=
g 30
force mass
=
2 y gA gA (60 A )
=
= (2 × 30 A) × 1 = 60 A yg 30
y = ω2 y
Hence acceleration is directly proportional to displacement, so the motion is simple harmonic motion. The time period T is given by T=
2π
ω
30 30 = 2π = 1.098 s g 9 8 0
= 2π
6. A cubical body (side 0.1 m and mass 0.002 kg) floats in water. It is pressed and then released so that it oscillates vertically. The time period is (A) 0.01 s
(B) 0.02 s
(C) 0.03 s
Ans (C) Ans (C) Force in displaced position = Mg − ρ (d + x) Ag But Mg = ρ d A g
∴ Force = ρ d A g − ρ (d + x) Ag = − ρ × A g The equation of motion is given by M 2
d x dt
2
=−
ρAg M
2IIT1617PPWS14S
d2 x dt
2
= −ρ A gx
x = −ω2 x
10
(D) 0.04 s
This is a equation of SHM
M 0.002 = 2π = 0.028 second. 2 ρ × × Ag 1 0 0 0 ( 0 . 1 ) 9 . 8
2π
T=
= 2π
ω
7. A copper wire is held at the two ends by rigid support. At 30 °C the wire is just taut, with negligible 11
−2
tension. The speed of transverse waves in this wire at 10 °C is (Y = 1.3 × 10 Nm , α = 1.7 × 10
−5
°C−1
−3
3
and ρ = 9 × 10 kg m ) (A) 70 ms
−1
−1
(B) 30 ms
(C) 90 ms
−1
−1
(D) 120 ms
Ans (A) Ans (A) T
v=
TL
=
µ
M
T
Thermal stress,
A
YAα∆θL
v=
where µ =
M
M L
=
Yα∆θ
ρ
1.3 × 1011 × 1.7 ×10 −5 × 20
v=
∆L ∵ α = L∆θ M ∵ µ = A = ρ × L
= yα∆θ
9 × 10
3
= 70 m s −1
8. A wave pulse starts propagating in +x direction along a non-uniform wire of length L under a tension T with mass per unit length given as µ = µ0 + αx, where µ0 and α are constants. The time taken by the pulse to travel from the lighter end (x = 0) to the heavier end is 2 2 ( −µ0 + αL)3/ 2 + µ30/ 2 (µ0 + αL)3/ 2 − µ30/ 2 (A) (B) 3α T 3α T 2 2 (µ0 + αL)3/ 2 + µ30 / 2 (µ0 + αL) 3/ 2 − µ30/ 2 (C) (D) 2α T 2α T Ans (B) Velocity of transverse wave in string, v = dx dt
=
T
µ 0 + αx
or
L
∫ (µ
T
µ
µo + αx dx = T dt L
t/2
0
+ αx ) dx dx = T
0
∫ dt 0
2(µ0 + αx )3/ 2 3α 2 3α t=
L t
= T t ]0 0
3/ 2 3/ 2 [µ0 + αL) − µ0 = T t
2 3α T
[(µ0 + αL)3 / 2 − µ0 3 / 2 ]
9. Figure shows three identical springs A, B, C. When a 4 kg weight is hung on A, it descends by 1 cm. When a 6 kg weight is hung on C, it will descend by (A) 1.5 cm (B) 3.0 cm (C) 4.5 cm 2IIT1617PPWS14S
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(D) 6.0 cm Ans (B) Ans (B) mg = kx kx ⇒ k =
mg x
=
4 × 10 1 × 10 10
−2
= 4000 Nm −1
When B and C are in series, k eff =
∴ x=
mg k eff
=
6 × 10 2000
k 2
= 2000 Nm −1
= 3 cm.
10. A simple pendulum of length l has a time period T for small oscillation. A fixed obstacle is placed directly below the point of suspension, so that only the lower quarter of the string continues oscillations. The pendulum is released from rest at a certain point O. The period of oscillation (assuming small angles) will be (A) T T (B) 4 3T (C) 4 T (D) 1+ 3 2
(
)
Ans (C) Ans (C) T′ =
1
l
+ 2π 2π 2 g
l
/4
1 T 3T = T + = g 2 2 4
11. The radius of steel wire A is twice that of B and the tension in A is half that in B. If transverse waves of same frequency are generated in these two steel wires, then the ratio of the velocities of waves in A and B is (A) 1 : 2
(B) 1 : 2
(C) 1 : 2 2
(D) 3 : 2 2
Ans (C) v= vA =
T
µ
T
=
πr 2 ρ
TA rA πρ
, vB =
TB rB πρ
⇒
vA vB
=
TA rB v . ⇒ A = TB rA vB
1 1 1 . = 2 2 2 2
12. Two sources A and B are sounding notes of frequency 680 Hz. A listener moves from A to B with a −1
constant velocity u. If the speed of sound is 340 m s , the value of u so that he hears 10 beats per second is (A) 2.0 m s
−1
(B) 2.5 m s
−1
(C) 3.0 m s
Ans (B) Ans (B)
v − u v v + u and moves towards, B ∴ f B′ = f v v + u v − u given f B′ − f A′ = f − f v v Listener moves away from A, ∴ f A′ = f
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−1
(D) 3.5 m s
−1
10 =
f v
[ 2u ] ⇒ u =
u = 2.5 m s
10 v 2f
=
10 × 340 2 × 680
−1 −1
13. A wire having a linear density 0.1 kg m is kept under a tension 490 N. It is observed that it resonates at a frequency of 400 Hz and the next higher frequency 450 Hz. The length of wire is (A) 0.4 m
(B) 0.7 m
(C) 0.6 m
(D) 0.49 m
Ans (B)
1
f = n
T
2l µ 1 T 400 = n 2l µ 1 T 450 = (n + 1) 2l µ ( 2) (1)
⇒
450 400
=
n +1 n
1
1 2l
… (2)
⇒ n =8
Using (1) 400 = n 400 = 8
… (1)
T
2l µ 490 ⇒ l = 0 .7 m 0.1
Read the passage given below and answer the following questions b choosing the correct alternative
A string of natural length 2 l can just support certain weight, when it is stretched till its whole length is 3 l. One end of the string is now attached to a point on a smooth horizontal table and the same weight is attached to the other end and can move on the table. When the weight is pulled out to any distance and let go, the string becomes slack again after some time. 14. If f is the force required to produce unit strain in the string, then weight required to increase the length from 2l to 3l is (A) f
(B)
f 3
(C)
f 2
(D)
f 4
Ans (C) Ans (C) If L is the natural length of the string and f is the force force required to produce unit strain in the string string then stretching force required to produce an extension of AL is f F = ∆L L l f f = 2l 2 15. The period of oscillation of the mass attached to the string is (A) T = 2π
l
g
(B) T = π
l
g
(C) T = 2π
Ans (A) Ans (A) 2IIT1617PPWS14S
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a 2g
(D) T = 2π
2a g
One end of the string is attached to a smooth table and the other end to a mass M. Let it be pulled through distance x. Force producing this stretching, F =
∴ T = 2π
m k
f L
mg x mgx ⇒ Force constant k = = l l 2l
x = 2 mg
l m = 2π mg g
= 2π
l
16. Time interval in which string becomes slack is (A) π
l
g
(B)
π
l
2
g
(C) 2π
l
g
(D)
π
l
2
2g
Ans (B) Ans (B)
T second after the 4
String will become slack when it reaches its equilibrium length, which happens at mass m is released from extreme position of the string. Hence time required =
T 4
=
1 4
2π
l
g
=
π
l
2
g
.
!PP 17. A mass m is suspended by mass of two coiled spring which have the same length in unscratched condition as in figure. Their constant are k 1 and k 2 respectively. When set into vertical vibrations. the period will be
m (A) 2π k1k 2
k (B) 2π m 1 k 2
m (C) 2π k1 − k 2
m (D) 2π k1 + k 2
Ans (D) Ans (D) Given spring system has parallel combination, so K eq = K1 + K 2 and time period T = 2π
m
( K1 + K 2 )
18. The equation of a damped simple harmonic motion is m
d2 x dt
2
+b
dx dt
+ kx = 0. Then the angular frequency
of oscillation is 1
2 k b 2 (A) ω = − m 4m 2 1
k b2 2 (C) ω = − m 4m 2IIT1617PPWS14S
1
b 2 k (B) ω = − m 4m 2
k b2 (D) ω = − m 4m 14
Ans (A) 19. The angular velocity and the amplitude of a simple pendulum is ‘ ω’ and ‘a’ respectively. At displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is (A)
X 2ω2 a 2 − X 2ω2
(B)
X2 a 2 − X2
(C)
( a 2 − X 2 ω2 ) X 2 ω2
(D)
(a 2 − X2 ) X2
Ans (D) Ans (D) 1
Kinetic energy T =
2
mω2 ( a 2 − x 2 ) and potential energy, V =
1 2
mω 2 x 2 ∴
T V
=
a2 − x2 x2
20. A particle is oscillating in SHM. What fraction of total energy is kinetic when the particle is at the mean position (A is the amplitude of oscillation) 3 2 4 (A) (B) (C) 4 4 7
(D)
A 2
from
5 7
Ans (A) Ans (A) K f =
K E at
1
A
K Total
2 = 2
mω2 A 2 − 1 2
⇒ K f =
A2
4
mω2 A 2
3 4
21. If ‘x’, ‘v’ and ‘a’ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then which of the following does not change with time aT aT aT (A) a 2T 2 + 4π2v 2 (B) (C) (D) x 2πv v Ans (B) 2
Acceleration a = ω x
∴
aT
=
x
ω2 xT x
2
4π 2 2π =ω T= T = T T 2
It is a constant term for S.H.M. i.e., it does not change with time. 22. The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4 / 3 s is (A) (B) (C)
3 32
−π
π2 c m / s 2 2
c m / s2
32
π2 32
(D) −
c m / s2 3
32
π2 c m / s 2
Ans (D) From given graph amplitude (a) = 1 cm 2IIT1617PPWS14S
15
Time period (T) = 8 s 2π π
∴
ω=
8
=
4 2
Acceleration A = − ω a sin ωt π2 4 π 4 at t = s, A = − ×1× sin × 3 16 4 3
⇒A=
−π2 16
− 3 2 π π cm / s2 ⇒A = 32 3
sin
Read the passage given below and answer the following questions b choosing the correct alternative
Two identical blocks P and Q have equal masses and are connected to two identical springs as shown. Initially the springs are unstretched. The block P A is moved to left by and Q by A to the right side from equilibrium 2 position. Both the blocks are released simultaneously and they undergo perfectly inelastic collision. In perfectly inelastic collision between two blocks, the momentum remains conserved and blocks stick together. Initially, time period of both the blocks was T. Angular frequency of spring block system is ω = maximum speed of particle in SHM is ωA, where A is the amplitude of oscillation. 23. The energy of oscillation of combined mass is T (A) T (B) 2
(C) 2T
(D)
T 2
Ans (A) Ans (A) Blocks collide at their mean position as time period is same for both. After collision combined mass is 2 m and k eff eff = 2 k, hence time period remains same. T = 2π 24. The amplitude of the combined mass is A (A) 3A (B) 2
(C)
2A 3
(D)
A 4
Ans (D) Ans (D) From conservation of linear momentum A ωA mω − mωA = (2m) v′ ⇒ v′ = 2 4 A ωA ⇒ Amplitude = . ∴ Velocity at mean position = 4 4 25. The energy of oscillation of combined mass is kA 2 2 (A) kA (B) 2
(C)
Ans (C) Ans (C) 2
mω2 A 2 kA 2 ωA = = E = ( 2 m ) v′ = ( 2 m ) 2 2 16 16 4 1
2IIT1617PPWS14S
2
1
16
kA 2 16
(D)
kA 2 8
m k
k m
and
Read the passage given below and answer questions b choosing the correct alternative
Longitudinal standing waves can be produced in columns of air in pipes. The closed end of the pipe is a node. The open end of a pipe is always an antinode. The lowest natural frequency is called fundamental frequency f 0 and has wavelength λ0. A harmonic is an integral multiple of fundamental frequency (nf) with n = 1, 2, 3 … For a complete cycle of a wave of wavelength λ, the distance between any two consecutive nodes or any two consecutive antinodes is half the wave length. i.e.
λ 2
(Assume v = 340 ms
−1
)
26. The auditory canal of the outer ear is 3 cm, closed at one end by the ear drum. The fundamental frequency associated with air column is 2
(A) 1.1 × 10 Hz
3
(B) 2.8 × 10 Hz
3
(C) 5.6 × 10 Hz
4
(D) 1.7 × 10 Hz
Ans (B) Ans (B) v 340 f 0 = = = 2.8 × 10 3 Hz −2 4L 4 × 3 × 10 27. A pipe resonates at 60 Hz, 100 Hz, and 140 Hz. How long is the pipe? (A) 1.4 m
(B) 2.8 m
(C) 4.3 m
(D) 8.5 m
Ans (C) Ans (C) The difference between successive overtones is constant (= 40 Hz)
⇒ fundamental frequency must be 20 Hz. 3(20 Hz) = 60 Hz, 5(20 Hz) 100 Hz, 7(20 Hz) 140 Hz. Overtones are odd multiples of the fundamental, hence it is a closed pipe. v v 340 = = 4.25 m ≈ 4.3 m f 0 = ,∴ l = 4L 4f 0 4 × 20 28. A pipe of length L is closed at both the ends. What is its fundamental wavelength? L (A) (B) L (C) 2L (D) 4L 2 Ans (C) Ans (C)
L=
λ0 2
⇒ λ 0 = 2L
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2IIT1617PPWS14S
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