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SIMPLE HARMONIC MOTION:--
EXERCISES Q-41 TO Q-50
41. Assume that a tunnel is dug across the earth (radius = R) passing through its center. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of √(gR) (b) It is released from a height R above the tunnel (c) it is thrown vertically upward along the length of the tunnel with a speed of √(gR).
ANSWER: (a) Force at the surface of the earth on the particle =mg. It will be in a simple harmonic motion. So,
Acceleration a = ⍵²x =mg/m =g
→⍵² = g/x =g/R at the end of the tunnel.
→⍵ = √(g/R)
Time period T =2π/⍵ =2π√(R/g)
Now, v =⍵√(A²-x²), at x=R, v=√(gR)
→√(gR) = √(g/R)√(A²-R²)
→R = √(A²-R²)
→R² =A² - R²
→A² =2R²
→A =R√2
The displace ment is given by
x = A.sin{√(g/R)t+ẟ}
At t = 0, x = R
→A.sinẟ=R
→sinẟ =R/A =R/R√2 =1/√2
→ẟ = π/4 or 3π/4
We shall take the greater value 3π/4 which is for the particle entering the tunnel. π/4 is for the imaginary value when the particle coming out of the tunnel.
Now, x = R√2.sin{√(g/R)t+3π/4}
Since we have taken center of the earth as origin, the displacement at other end of the tunnel x =-R
So, -R = R√2.sin{√(g/R)t+3π/4}
→sin{√(g/R)t+3π/4}=-1/√2 = sin(5π/4)
→ √(g/R)t+3π/4 = 5π/4
→√(g/R)t=2π/4 = π/2
→t = (π/2)√(R/g)
(b) When the particle is released from a height R above the tunnel, its velocity at the entrance to the tunnel v is given by,
½mv² = GMm/(R+R) {Equating P.E. at height R to K.E. at the entrance}
→v² =2GM/2R =GM/R
→v² =gR²/R {Since GM=gR²}
→v² =gR
→v =√(gR)
So the velocity of the particle at the entry is √(gR) which is the same in the above problem (a). Hence the time taken to cover the tunnel is the same = (π/2)√(R/g)
(c) When the particle is thrown upward along the tunnel with a velocity √(gR) its velocity at the entry to the tunnel will also be √(gR) because P.E. at the entry level will be the same hence the K.E. (and hence the velocity) will also be the same. Now the case becomes the same as in above problem (a). So the same time will be taken = (π/2)√(R/g)
42. Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R/2 from the earth's center where R is the radius of the earth. The wall of the tunnel is frictionless. (a) Find the gravitational force exerted by the earth on a particle of mass m placed in the tunnel at a distance x from the center of the tunnel. (b) Find the component of this force along the tunnel and the perpendicular to the tunnel. (c) Find the normal force exerted by the wall on the particle. (d) Find the resultant force on the particle. (e) Show that the motion of the particle in the tunnel is simple harmonic and find the time period.
ANSWER: Let us draw a figure as below where the particle of mass m is at A. The center of the tunnel is C. Given CA =x and OC=R/2.
The figure for Q-42
The distance of the particle from the center of the earth OA =d,
=√{OC²+CA²}
=√{R²/4+x²}
The magnitude of the gravitational intensity at this point E=GM'/d²
Where the M' is the mass of the earth portion having radius d. If ρ is the density of the earth, then M' =4πd³ρ/3 and M =4πR³ρ/3.
So, M'/M =d³/R³
→M' =Md³/R³
Now E =GMd³/R³d² =GMd/R³
The gravitational force on the particle m at this point by the earth F
=Em
=GMmd/R³
=(GMm/R³)√{R²/4+x²}
(b) Let θ be the angle between AO and AC. The component of this force along the tunnel =F.cosθ =Fx/d =Fx/√(R²/4+x²)
=(GMm/R³)√{R²/4+x²}{x/√(R²/4+x²)}
=GMmx/R³
The component of this force perpendicular to the tunnel =F.sinθ
=F.(R/2)/d
=(GMm/R³)√{R²/4+x²}(R/2)/√{R²/4+x²}
=GMm/2R²
(c) The normal force exerted by the tunnel wall on the particle will be equal and opposite to the magnitude of the component of force F perpendicular to the wall. From the solution above (b),
Normal force =F.sinθ =GMm/2R²
(d) The component of the force perpendicular to the tunnel wall F.sinθ is balanced by the normal force. Since the tunnel walls are smooth there is no frictional force along the tunnel. Hence the net/resultant force on the particle is the component of the force along the tunnel F.cosθ =GMmx/R³
(e) The force along the tunnel (towards the movement)
Fₓ = F.cosθ =GMmx/R³ =(GMm/R³)x =m⍵²x
Since the force is proportional to the displacement, it is a simple harmonic motion with ⍵ =√(GM/R³)
The time period T =2π/⍵ =2π√(R³/GM)
43. A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with an acceleration a₀ (b) is going down with an acceleration a₀ and (c) is moving with a uniform velocity.
ANSWER: (a) When the elevator is going up, if we observe the bob from the non-inertial frame of the lift, additional pseudo force ma₀ is applied opposite to the direction of acceleration. Total force on the bob =mg+ma₀ =m(g+a₀). Therefore the acceleration of the bob
g' =m(g+a₀)/m = g+a₀
Hence the time period =2π√(l/g') =2π√{l/(g+a₀)}.
(b) Similarly, when the elevator is going down with an acceleration a₀, the pseudo force ma₀ is upwards. The force on the bob =m(g-a₀).
Hence the acceleration g' = m(g-a₀)/m =g-a₀
Hence the time period =2π√(l/g')
=2π√{l/(g-a₀)}
(c) Since the elevator is moving with a uniform velocity, there is no acceleration of the elevator. The elevator frame is inertial frame and the weight (force on the bob) remains mg. Hence the acceleration of the bob = mg/m =g.
So, the time period T = 2π√(l/g)
44. A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes π/3 seconds to complete one oscillation. Find the acceleration of the elevator.
ANSWER: Given T = π/3 s, l =1 ft in FPS system g =32 ft/s²
Let the acceleration of the elevator =a. If we assume its direction upwards. With the application of pseudo force in downward direction total force on the bob =mg+ma =m(g+a)
Hence the acceleration g'= m(g+a)/m =g+a
Time period T =2π√(l/g')
→π/3 =2π√(1/g')
→√g' =2π/(π/3) =6
g' = 36 ft/s²
→g+a = 36
→a = 36-g =36-32 =4 ft/s²
Since it is positive so the direction assumed is correct i.e. upward.
45. A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.
ANSWER: Let the length of the simple pendulum = l. T =4 s.
So, T =2π√(l/g)
→4 = 2π√(1/g)
→√(l/g) =2/π
→l =4g/π²
When the car is moving with an acceleration a, to analyze the motions from the non-inertial frame of the car we apply a pseudo force mg in the opposite direction of the acceleration to the bob. Now the bob has an acceleration g vertically downwards and another acceleration a horizontally backward. So the magnitude of the net acceleration of the bob g' =√(g²+a²)
The time period now = 2π√(l/g') = 3.99
→2π√(4g/π²g') =3.99
→4√(g/g') =3.99
→g/g' =(3.99/4)²
→g' = g(4/3.99)²
→√(g²+a²) =g(4/3.99)²
→g²+a² = g²(4/3.99)⁴
→a² =g²{(4/3.99)⁴-1} =g²{(4⁴-3.99⁴)/3.99⁴}
→a =g√{2.55/253.45} ≈g√0.01
→a = 0.1g =g/10
46. A simple pendulum of length l is suspended from the ceiling of a car moving with a speed v on a circular horizontal road of radius r. (a) Find the tension in the string when it is at rest with respect to the car. (b) Find the time period of small oscillation.
ANSWER: (a) When the car is at rest with respect to the car, there will be three forces acting on the bob. The weight mg, the centrifugal force mv²/r and the tension in the string (say) P.
Figure for Q-46
Since the three forces are in equilibrium, the tension in the string will be equal and opposite to the resultant of the weight mg and the centrifugal force mv²/r.
So the magnitude of the tension in the string = P
=√[(mg)²+(mv²/r)²]
=m[g²+v⁴/r²]½
=ma
Where a = [g²+v⁴/r²]½
(b) The time period of the small oscillations = T =2π√(l/a)
→T = 2π√[l/{g²+v⁴/r²}½]
47. The earring of a lady shown in the figure (12-E18) has a three cm long light suspension wire. (a) Find the time period of small oscillations if the lady is standing on the ground. (b) The lady now sits in a merry-go-round moving at 4 m/s in a circle of radius 2m. Find the time period of small oscillations of the ear-ring.
Figure for Q-47
ANSWER: Given l = 3 cm =0.03 m.
(a) When the lady is standing on the ground, the time period of small oscillations = 2π√(l/g) =2π√(0.03/9.8) =2π√0.0031=2π*0.06
=0.34 s
(b) When the lady is in a merry go round the velocity v = 4 m/s, r = 2 m. The centrifugal force on the bob of the ring = mv²/r (horizontal) and the weight of the bob = mg. The net force on the bob =[(mg)²+(mv²/r)²]½
=m [g²+(v²/r)²]½
The acceleration a = m [g²+(v²/r)²]½ /m = [g²+(v²/r)²]½
=√[9.8²+(16/2)²]
=√(96.04+64)
=√160.04
=12.65 m/s²
The time period = 2π√(0.03/12.65)
=2*3.14*0.048
=0.30 s
48. Find the time period of small oscillations of the following systems. (a) A meterstick suspended through the 20 cm mark. (b) A ring of mass m and radius r suspended through a point on its periphery. (c) A uniform square plate of edge a suspended through a corner. (d) A uniform disc of mass m and radius r suspended through a point r/2 away from the center.
ANSWER: (a) When the meter stick is suspended through 20 cm mark it is a physical pendulum. Its time period for small oscillations is given as,
T = 2π√(I/mgl)
l = distance between the point of support and the center of mass
=30 cm {CoM will be at 50 cm}
=0.30 m
M.I. about the horizontal axis through the point of support =M.I. about the horizontal axis through CoM + ml²
=mL²/12 + ml²
=m(1/12+0.30²) {Since L = 1}
=m(0.083+0.09)
=0.173m kg-m².
Thus T = 2π√{0.173m/(m*9.8*0.30)}
=2π√(0.173/9.8*0.3)
=2π√0.058
=1.51 s
(b) CoM of the ring is at the center of the ring. The point of suspension is at the periphery of the ring. Hence l = r.
M.I. of the ring about a horizontal axis through the center = mr²
M.I. of the ring about a horizontal axis through the support =mr²+mr²
=2mr²
The time period = 2π√(I/mgl)
= 2π√(2mr²/mgr)
2π√(2r/g)
(c) For the square plate hanged through a corner,
l = a/√2
M.I. about a horizontal axis through the point of suspension
= ma²/6 + m(a/√2)²
=m{a²/6+a²/2}
=2ma²/3
Hence the time period =2π√{(2ma²/3)/(mga/√2)}
=2π√{2√2a/3g}
=2π√{√8a/3g}
(d) In the case of the uniform disc suspended through a point at a distance r/2 away from the center, l = r/2.
M.I. about the horizontal axis through the point of suspension =½mr²+m(r/2)²
=½mr²+¼mr²
=¾mr²
The time period =2π√{(¾mr²)/(mgr/2)}
=2π√(3r/2g)
49. A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.
ANSWER: The rod is a physical pendulum. For small oscillations time period = 2π√{I/(mgl')}
l' = l/2
M.I. about the end = ml²/12 +m(l/2)²
=ml²(1/12 + ¼)
=ml²(4/12)
=ml²/3
The time period = 2π√{(ml²/3)/(mgl/2)}
=2π√(2l/3g)
If the length of a simple pendulum having the same time period is L, then
T=2π√(L/g) = 2π√(2l/3g)
→L/g = 2l/3g
→L = 2l/3
50. A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillation. What should be the distance of the hole from the center for it to have the minimum time period?
ANSWER: Let the distance from the CoM to the point of suspension be x.
M.I. of the disc about the point of suspension = ½mr² + mx²
=m(r²/2+x²)
Time period T= 2π√(I/mgx)
=2π√{m(r²/2+x²)/mgx}
=2π√{(r²/2+x²)/gx}
→T² = 4π²r²/2gx + 4π²x/g -------- (i)
For x to be minimum dT/dx = 0
Differentiating (i) w.r.t. x
2T*dT/dx = -4π²r²/2gx² +4π²/g
dT/dx = {-4π²r²/2gx² +4π²/g}/2T =0
→-4π²r²/2gx² +4π²/g =0
→-r²/2gx² + 1/g = 0
→1 = r²/2x²
→2x² = r²
→x² =r²/2
→x =r/√2
So for the minimum time period the distance of the hole from the center should be r/√2.
Putting the value of x in (i)
T² = 4π²r²/(2gr/√2) + 4π²r/√2g
→T² = 4π²r/√2g + 4π²r/√2g
→T² = 8π²r/√2g =4√2π²r/g
→T = 2π√(r√2/g)
It is the minimum time period.
41. Assume that a tunnel is dug across the earth (radius = R) passing through its center. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of √(gR) (b) It is released from a height R above the tunnel (c) it is thrown vertically upward along the length of the tunnel with a speed of √(gR).
ANSWER: (a) Force at the surface of the earth on the particle =mg. It will be in a simple harmonic motion. So,
Acceleration a = ⍵²x =mg/m =g
→⍵² = g/x =g/R at the end of the tunnel.
→⍵ = √(g/R)
Time period T =2π/⍵ =2π√(R/g)
Now, v =⍵√(A²-x²), at x=R, v=√(gR)
→√(gR) = √(g/R)√(A²-R²)
→R = √(A²-R²)
→R² =A² - R²
→A² =2R²
→A =R√2
The displace ment is given by
x = A.sin{√(g/R)t+ẟ}
At t = 0, x = R
→A.sinẟ=R
→sinẟ =R/A =R/R√2 =1/√2
→ẟ = π/4 or 3π/4
We shall take the greater value 3π/4 which is for the particle entering the tunnel. π/4 is for the imaginary value when the particle coming out of the tunnel.
Now, x = R√2.sin{√(g/R)t+3π/4}
Since we have taken center of the earth as origin, the displacement at other end of the tunnel x =-R
So, -R = R√2.sin{√(g/R)t+3π/4}
→sin{√(g/R)t+3π/4}=-1/√2 = sin(5π/4)
→ √(g/R)t+3π/4 = 5π/4
→√(g/R)t=2π/4 = π/2
→t = (π/2)√(R/g)
(b) When the particle is released from a height R above the tunnel, its velocity at the entrance to the tunnel v is given by,
½mv² = GMm/(R+R) {Equating P.E. at height R to K.E. at the entrance}
→v² =2GM/2R =GM/R
→v² =gR²/R {Since GM=gR²}
→v² =gR
→v =√(gR)
So the velocity of the particle at the entry is √(gR) which is the same in the above problem (a). Hence the time taken to cover the tunnel is the same = (π/2)√(R/g)42. Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R/2 from the earth's center where R is the radius of the earth. The wall of the tunnel is frictionless. (a) Find the gravitational force exerted by the earth on a particle of mass m placed in the tunnel at a distance x from the center of the tunnel. (b) Find the component of this force along the tunnel and the perpendicular to the tunnel. (c) Find the normal force exerted by the wall on the particle. (d) Find the resultant force on the particle. (e) Show that the motion of the particle in the tunnel is simple harmonic and find the time period.
ANSWER: Let us draw a figure as below where the particle of mass m is at A. The center of the tunnel is C. Given CA =x and OC=R/2.
 |
| The figure for Q-42 |
The distance of the particle from the center of the earth OA =d,
=√{OC²+CA²}
=√{R²/4+x²}
The magnitude of the gravitational intensity at this point E=GM'/d²
Where the M' is the mass of the earth portion having radius d. If ρ is the density of the earth, then M' =4πd³ρ/3 and M =4πR³ρ/3.
So, M'/M =d³/R³
→M' =Md³/R³
Now E =GMd³/R³d² =GMd/R³
The gravitational force on the particle m at this point by the earth F
=Em
=GMmd/R³
=(GMm/R³)√{R²/4+x²}
(b) Let θ be the angle between AO and AC. The component of this force along the tunnel =F.cosθ =Fx/d =Fx/√(R²/4+x²)
=(GMm/R³)√{R²/4+x²}{x/√(R²/4+x²)}
=GMmx/R³
The component of this force perpendicular to the tunnel =F.sinθ
=F.(R/2)/d
=(GMm/R³)√{R²/4+x²}(R/2)/√{R²/4+x²}
=GMm/2R²
(c) The normal force exerted by the tunnel wall on the particle will be equal and opposite to the magnitude of the component of force F perpendicular to the wall. From the solution above (b),
Normal force =F.sinθ =GMm/2R²
(d) The component of the force perpendicular to the tunnel wall F.sinθ is balanced by the normal force. Since the tunnel walls are smooth there is no frictional force along the tunnel. Hence the net/resultant force on the particle is the component of the force along the tunnel F.cosθ =GMmx/R³
(e) The force along the tunnel (towards the movement)
Fₓ = F.cosθ =GMmx/R³ =(GMm/R³)x =m⍵²x
Since the force is proportional to the displacement, it is a simple harmonic motion with ⍵ =√(GM/R³)
The time period T =2π/⍵ =2π√(R³/GM)
43. A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with an acceleration a₀ (b) is going down with an acceleration a₀ and (c) is moving with a uniform velocity.
ANSWER: (a) When the elevator is going up, if we observe the bob from the non-inertial frame of the lift, additional pseudo force ma₀ is applied opposite to the direction of acceleration. Total force on the bob =mg+ma₀ =m(g+a₀). Therefore the acceleration of the bob
g' =m(g+a₀)/m = g+a₀
Hence the time period =2π√(l/g') =2π√{l/(g+a₀)}.
(b) Similarly, when the elevator is going down with an acceleration a₀, the pseudo force ma₀ is upwards. The force on the bob =m(g-a₀).
Hence the acceleration g' = m(g-a₀)/m =g-a₀
Hence the time period =2π√(l/g')
=2π√{l/(g-a₀)}
(c) Since the elevator is moving with a uniform velocity, there is no acceleration of the elevator. The elevator frame is inertial frame and the weight (force on the bob) remains mg. Hence the acceleration of the bob = mg/m =g.
So, the time period T = 2π√(l/g)
44. A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes π/3 seconds to complete one oscillation. Find the acceleration of the elevator.
ANSWER: Given T = π/3 s, l =1 ft in FPS system g =32 ft/s²
Let the acceleration of the elevator =a. If we assume its direction upwards. With the application of pseudo force in downward direction total force on the bob =mg+ma =m(g+a)
Hence the acceleration g'= m(g+a)/m =g+a
Time period T =2π√(l/g')
→π/3 =2π√(1/g')
→√g' =2π/(π/3) =6
g' = 36 ft/s²
→g+a = 36
→a = 36-g =36-32 =4 ft/s²
Since it is positive so the direction assumed is correct i.e. upward.
45. A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.
ANSWER: Let the length of the simple pendulum = l. T =4 s.
So, T =2π√(l/g)
→4 = 2π√(1/g)
→√(l/g) =2/π
→l =4g/π²
When the car is moving with an acceleration a, to analyze the motions from the non-inertial frame of the car we apply a pseudo force mg in the opposite direction of the acceleration to the bob. Now the bob has an acceleration g vertically downwards and another acceleration a horizontally backward. So the magnitude of the net acceleration of the bob g' =√(g²+a²)
The time period now = 2π√(l/g') = 3.99
→2π√(4g/π²g') =3.99
→4√(g/g') =3.99
→g/g' =(3.99/4)²
→g' = g(4/3.99)²
→√(g²+a²) =g(4/3.99)²
→g²+a² = g²(4/3.99)⁴
→a² =g²{(4/3.99)⁴-1} =g²{(4⁴-3.99⁴)/3.99⁴}
→a =g√{2.55/253.45} ≈g√0.01
→a = 0.1g =g/10
46. A simple pendulum of length l is suspended from the ceiling of a car moving with a speed v on a circular horizontal road of radius r. (a) Find the tension in the string when it is at rest with respect to the car. (b) Find the time period of small oscillation.
ANSWER: (a) When the car is at rest with respect to the car, there will be three forces acting on the bob. The weight mg, the centrifugal force mv²/r and the tension in the string (say) P.
 |
| Figure for Q-46 |
Since the three forces are in equilibrium, the tension in the string will be equal and opposite to the resultant of the weight mg and the centrifugal force mv²/r.
So the magnitude of the tension in the string = P
=√[(mg)²+(mv²/r)²]
=m[g²+v⁴/r²]½
=ma
Where a = [g²+v⁴/r²]½
(b) The time period of the small oscillations = T =2π√(l/a)
→T = 2π√[l/{g²+v⁴/r²}½]
47. The earring of a lady shown in the figure (12-E18) has a three cm long light suspension wire. (a) Find the time period of small oscillations if the lady is standing on the ground. (b) The lady now sits in a merry-go-round moving at 4 m/s in a circle of radius 2m. Find the time period of small oscillations of the ear-ring.
|
| Figure for Q-47 |
ANSWER: Given l = 3 cm =0.03 m.
(a) When the lady is standing on the ground, the time period of small oscillations = 2π√(l/g) =2π√(0.03/9.8) =2π√0.0031=2π*0.06
=0.34 s
(b) When the lady is in a merry go round the velocity v = 4 m/s, r = 2 m. The centrifugal force on the bob of the ring = mv²/r (horizontal) and the weight of the bob = mg. The net force on the bob =[(mg)²+(mv²/r)²]½
=m [g²+(v²/r)²]½
The acceleration a = m [g²+(v²/r)²]½ /m = [g²+(v²/r)²]½
=√[9.8²+(16/2)²]
=√(96.04+64)
=√160.04
=12.65 m/s²
The time period = 2π√(0.03/12.65)
=2*3.14*0.048
=0.30 sANSWER: (a) When the meter stick is suspended through 20 cm mark it is a physical pendulum. Its time period for small oscillations is given as,
T = 2π√(I/mgl)
l = distance between the point of support and the center of mass
=30 cm {CoM will be at 50 cm}
=0.30 m
M.I. about the horizontal axis through the point of support =M.I. about the horizontal axis through CoM + ml²
=mL²/12 + ml²
=m(1/12+0.30²) {Since L = 1}
=m(0.083+0.09)
=0.173m kg-m².
Thus T = 2π√{0.173m/(m*9.8*0.30)}
=2π√(0.173/9.8*0.3)
=2π√0.058
=1.51 s
(b) CoM of the ring is at the center of the ring. The point of suspension is at the periphery of the ring. Hence l = r.
M.I. of the ring about a horizontal axis through the center = mr²
M.I. of the ring about a horizontal axis through the support =mr²+mr²
=2mr²
The time period = 2π√(I/mgl)
= 2π√(2mr²/mgr)
2π√(2r/g)
(c) For the square plate hanged through a corner,
l = a/√2
M.I. about a horizontal axis through the point of suspension
= ma²/6 + m(a/√2)²
=m{a²/6+a²/2}
=2ma²/3
Hence the time period =2π√{(2ma²/3)/(mga/√2)}
=2π√{2√2a/3g}
=2π√{√8a/3g}
(d) In the case of the uniform disc suspended through a point at a distance r/2 away from the center, l = r/2.
M.I. about the horizontal axis through the point of suspension =½mr²+m(r/2)²
=½mr²+¼mr²
=¾mr²
The time period =2π√{(¾mr²)/(mgr/2)}
=2π√(3r/2g)
49. A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.
ANSWER: The rod is a physical pendulum. For small oscillations time period = 2π√{I/(mgl')}
l' = l/2
M.I. about the end = ml²/12 +m(l/2)²
=ml²(1/12 + ¼)
=ml²(4/12)
=ml²/3
The time period = 2π√{(ml²/3)/(mgl/2)}
=2π√(2l/3g)
If the length of a simple pendulum having the same time period is L, then
T=2π√(L/g) = 2π√(2l/3g)
→L/g = 2l/3g
→L = 2l/3
50. A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillation. What should be the distance of the hole from the center for it to have the minimum time period?
ANSWER: Let the distance from the CoM to the point of suspension be x.
M.I. of the disc about the point of suspension = ½mr² + mx²
=m(r²/2+x²)
Time period T= 2π√(I/mgx)
=2π√{m(r²/2+x²)/mgx}
=2π√{(r²/2+x²)/gx}
→T² = 4π²r²/2gx + 4π²x/g -------- (i)
For x to be minimum dT/dx = 0
Differentiating (i) w.r.t. x
2T*dT/dx = -4π²r²/2gx² +4π²/g
dT/dx = {-4π²r²/2gx² +4π²/g}/2T =0
→-4π²r²/2gx² +4π²/g =0
→-r²/2gx² + 1/g = 0
→1 = r²/2x²
→2x² = r²
→x² =r²/2
→x =r/√2
So for the minimum time period the distance of the hole from the center should be r/√2.
Putting the value of x in (i)
T² = 4π²r²/(2gr/√2) + 4π²r/√2g
→T² = 4π²r/√2g + 4π²r/√2g
→T² = 8π²r/√2g =4√2π²r/g
→T = 2π√(r√2/g)
It is the minimum time period.
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
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HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
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HC Verma's Concepts of Physics, Chapter-4, The Forces
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