Question

A Spring having with a spring constant 1200 N m⁻¹ is mounted on a horizontal table as shown in the figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance 2.0 cm and released. Determine(i) the frequency of oscillations(ii) maximum acceleration of the mass and (iii) the maximum speed of the mass.

Answer 1

Here the given body undergoes SHM

therefore we Time period=$\frac{2\pi }{w}$

and w=$\sqrt{\frac{k}{m} }$

therefore time perioid = $\frac{2\pi \sqrt{m} }{\sqrt{k} }$

therefore here m=3kg

k=1200

time period=$(2*3.14*\sqrt{3} )/\sqrt{1200}$     _____(1)

but frequcency =1/time period

therefore from 1

frequceny=3.18 sec^-1

Maximum acceleration

$a=w^{2}[\tex]   * amplitude of displacement</p><p>a=[tex]\frac{k}{m}$  *A

a=1200/3*0.02

a=8 m/sec^2

Maximum speed= A*w

=$0.02*\frac{\sqrt{1200} }{\sqrt{3} }$

=0.4 m/sec

Answer 2

Hii dear,

◆ Answer-
a) f = 3.18 Hz
b) a = 8 m/s^2
c) v = 0.4 m/s

◆ Explaination-
# Given-
m = 3 kg
k = 1200 N/m
A = 2 cm = 0.02 m

# Solution-
Angular velocity-
ω = √(k/m)
ω = √(1200/3)
ω = 20 rad/s

a) Frequency of oscillation is given by-
f = ω/2π
f = 20/(2×3.14)
f = 3.18 Hz

b) Maximum acceleration is given by-
a = Aω^2
a = 0.02×(20)^2
a = 8 m/s^2

c) Maximum velocity is given by-
v = aω
v = 0.02×20
v = 0.4 m/s

Hope this is helpful...