A 800 g body vibrates S.H.M with amplitude 0.30 m. The restoring force is 60 N and the displacement is 0.30 m. Find out (1) T (ii) `\vec a` (iii) `\vec v` (iv) K.E (v) P.E when the displacement is 12cm. (i) T = 1.3 s (ii) `\vec a` = 3 m s⁻² (iii) `\vec v` = 1.4 m s `\vec v` (iv) K.E = 7.6 J (v) PE = 1.44 J)
Given:
Mass of the body = m = 800 g = 0.8 kg
Amplitude = x₀ = 0.30 m
Force (restoring) = `\F_r` = 60 N
Displacement = x = 0.30 m
Given:
Mass of the body = m = 800 g = 0.8 kg
Amplitude = x₀ = 0.30 m
Force (restoring) = `\F_r` = 60 N
Displacement = x = 0.30 m
To Find:
(1) Time Period = T = ? (ii) acceleration = `\vec a` = ? when the displacement = x = 12cm = 0.12 m(iii) speed = `\vec v` = ? (iv) K.E = ? when the displacement = x = 12cm = 0.12 m(v) P.E = ? when the displacement = x = 12cm = 0.12 m
(1) Time Period = T = ?
(ii) acceleration = `\vec a` = ? when the displacement = x = 12cm = 0.12 m
(iii) speed = `\vec v` = ?
(iv) K.E = ? when the displacement = x = 12cm = 0.12 m
(v) P.E = ? when the displacement = x = 12cm = 0.12 m
Solution:
(1) Time Period = T = ? by using x = 0.3 m
The general formula for Time Period of a mass attached to spring is
T = 2Ï€ `\sqrt frac{m}{k}`
Here we have find first the value of spring constant due to mas of the block m₁. So by using Hook's Law
`\F_r` = k x
or
k = `\frac {F_r}{x}`
by putting values
k = `\frac {60 N}{0.30 m}`
k = 200 N m⁻¹
Now the Time period is
T = 2Ï€ `\sqrt frac{m}{k}`
by putting values
T = 2Ï€ `\sqrt frac{0.80 kg}{200 N m⁻¹}`
T = 2 ｘ 3.1416 ｘ `\sqrt {0.004 s²}`
T = 6.283 ｘ 0.063 s
T = 0.397 s
or
T = 0.4 s ------------------Ans.(i)
( NOTE: The Answer shown in the book T = 1.3 s can be only obtained if we use `\F_r` = 6 N thus obtaining K = 20 Nm⁻¹ ).
(ii) acceleration = `\vec a` = ? when the displacement = x = 12cm = 0.12 m
According to the Hook's Law
F = k x
and Newton Second law
F = m`\vec a`
by comparing we get
m`\vec a` = kx
or
`\vec a` = `\frac {kx}{m}`
by putting values
`\vec a` = `\frac {200 N m⁻¹ ｘ 0.12 m}{0.8 kg}`
`\vec a` = 30 m s⁻² ---------------Ans(ii)
( NOTE: The Answer shown in the book a = 3.0 s can be only obtained if we use `\F_r` = 6 N thus obtaining K = 20 Nm⁻¹ ).
(iii) speed = `\vec v` = ?
We know that
v = à´§ `\sqrt {x₀^2 - x^2}`
where à´§ = `\frac {2Ï€}{T}`
by putting values
ധ = `\frac {2 ｘ 3.1416}{1.3 s}`
à´§ = 4.8 rad s⁻¹ ----------------Ans(iii)
so,
v = à´§ `\sqrt {x₀^2 - x^2}`
by putting values
v = 4.8 rad s⁻¹ ｘ `\sqrt {(0.30m)^2 - (0.12 m)^2}`
v = 4.8 rad s⁻¹ ｘ `\sqrt {0.0756 m^2}`
v = 4.8 rad s⁻¹ ｘ 0.275
v = 1.319 m s⁻¹ ----------------------Ans. (iii)
(iv) K.E = ? when the displacement = x = 12cm = 0.12 m
The formula for K.E. is
K.E = `\frac{1}{2}` k `(x₀^2 - x^2)`
by putting values
K.E = `\frac{1}{2}`ｘ 200 N m⁻¹ï½˜`\((0.30m)^2 - (0.12 m)^2)`
K.E = 100 N m⁻¹ ｘ `\0.0756 m^2`
K.E = 7.56 J -----------------------Ans (iv)
(v) P.E = ? when the displacement = x = 12cm = 0.12 m
P.E = `\frac{1}{2}` k `\x^2`
putting values
P.E = `\frac{1}{2}` 200 N m⁻¹ `\(0.12 m)^2`
P.E = 100 N m⁻¹ ｘ 0.0144 m²
P.E = 1.44 J
************************************
(1) Time Period = T = ? by using x = 0.3 m
The general formula for Time Period of a mass attached to spring is
T = 2Ï€ `\sqrt frac{m}{k}`
Here we have find first the value of spring constant due to mas of the block m₁. So by using Hook's Law
`\F_r` = k x
or
k = `\frac {F_r}{x}`
by putting values
k = `\frac {60 N}{0.30 m}`
k = 200 N m⁻¹
Now the Time period is
T = 2Ï€ `\sqrt frac{m}{k}`
by putting values
T = 2Ï€ `\sqrt frac{0.80 kg}{200 N m⁻¹}`
T = 2 ｘ 3.1416 ｘ `\sqrt {0.004 s²}`
T = 6.283 ｘ 0.063 s
T = 0.397 s
or
T = 0.4 s ------------------Ans.(i)
( NOTE: The Answer shown in the book T = 1.3 s can be only obtained if we use `\F_r` = 6 N thus obtaining K = 20 Nm⁻¹ ).
(ii) acceleration = `\vec a` = ? when the displacement = x = 12cm = 0.12 m
According to the Hook's Law
F = k x
and Newton Second law
by comparing we get
m`\vec a` = kx
or
`\vec a` = `\frac {kx}{m}`
by putting values
`\vec a` = `\frac {200 N m⁻¹ ｘ 0.12 m}{0.8 kg}`
`\vec a` = 30 m s⁻² ---------------Ans(ii)
( NOTE: The Answer shown in the book a = 3.0 s can be only obtained if we use `\F_r` = 6 N thus obtaining K = 20 Nm⁻¹ ).
(iii) speed = `\vec v` = ?
We know that
v = à´§ `\sqrt {x₀^2 - x^2}`
where à´§ = `\frac {2Ï€}{T}`
by putting values
ധ = `\frac {2 ｘ 3.1416}{1.3 s}`
à´§ = 4.8 rad s⁻¹ ----------------Ans(iii)
so,
v = à´§ `\sqrt {x₀^2 - x^2}`
by putting values
v = 4.8 rad s⁻¹ ｘ `\sqrt {(0.30m)^2 - (0.12 m)^2}`
v = 4.8 rad s⁻¹ ｘ `\sqrt {0.0756 m^2}`
v = 4.8 rad s⁻¹ ｘ 0.275
v = 1.319 m s⁻¹ ----------------------Ans. (iii)
(iv) K.E = ? when the displacement = x = 12cm = 0.12 m
The formula for K.E. is
K.E = `\frac{1}{2}` k `(x₀^2 - x^2)`
by putting values
K.E = `\frac{1}{2}`ｘ 200 N m⁻¹ï½˜`\((0.30m)^2 - (0.12 m)^2)`
K.E = 100 N m⁻¹ ｘ `\0.0756 m^2`
K.E = 7.56 J -----------------------Ans (iv)
(v) P.E = ? when the displacement = x = 12cm = 0.12 m
P.E = `\frac{1}{2}` k `\x^2`
putting values
P.E = `\frac{1}{2}` 200 N m⁻¹ `\(0.12 m)^2`
P.E = 100 N m⁻¹ ｘ 0.0144 m²
P.E = 1.44 J
************************************
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